Using '|' from Bash's if construct prevents variable being set [duplicate] - bash

This question already has answers here:
bash: piping output from a loop seems to change the scope within the loop - why?
(2 answers)
Closed 6 years ago.
I've read that you can redirect I/O into and out of various constructs in Bash (such as "if" and "while"). When I was trying this out, I noticed that using a "|" at the end of an if construct prevented any variables that I overwrote inside the if construct from taking effect; but if I use ">" instead, then the variable modifications take effect.
#!/bin/bash
VAR_0='Unmodified'
if true
then
VAR_0='Changed'
fi | cat
echo "VAR_0: $VAR_0"
###########
VAR_1='Unmodified'
if true
then
VAR_1='Changed'
fi > tmpFile
rm tmpFile
echo "VAR_1: $VAR_1"
Running Bash version 4.3.11 on 64-bit Linux produces the following output:
VAR_0: Unmodified
VAR_1: Changed
Note the only difference is how I'm redirecting stdout from the if construct. Why is the "|" preventing VAR_0 from being changed?

TL;DR
Check out Bash FAQ 24.
It answers your question along with provided examples!
You're creating a SubShell which in turn being discarded along with its variables copy of the parent shell.
Take the following common mistake as an example:
var=0
some-command | while read -r line; do
printf 'var value inside subshell is: %s\n' "$((++var))"
done
printf 'But in the parent shell it stays: %s\n' "$var"
Output:
var value inside subshell is: 1
But in the parent shell it stays: 0
When this behvior is not desired, the solution is often to use Process Substitution to keep the varaibles updated. So, in our case ( or as can be seen generally in Bash FAQ 24 ) we do the following:
var=0
while read -r line; do
printf 'var value inside subshell is: %s\n' "$((++var))"
done < <(some-command)
printf 'And it stays the same: %s\n' "$var"
Output:
var value inside subshell is: 1
And it stays the same: 1

This occurs because pipelines use subshells; by piping to cat, you've made the first if-then block execute in a subshell. Hence, VAR_0='Changed' is changed, but only in the subshell.
Try:
#!/bin/bash
VAR_0='Unmodified'
if true
then
VAR_0='Changed'
echo "VAR_0: $VAR_0"
fi | cat
echo "VAR_0: $VAR_0"
Redirecting to a file (">") does not create a subshell; thus that variable persists.

Related

What shellenv command does? [duplicate]

After reading the Bash man pages and with respect to this post, I am still having trouble understanding what exactly the eval command does and which would be its typical uses.
For example, if we do:
$ set -- one two three # Sets $1 $2 $3
$ echo $1
one
$ n=1
$ echo ${$n} ## First attempt to echo $1 using brackets fails
bash: ${$n}: bad substitution
$ echo $($n) ## Second attempt to echo $1 using parentheses fails
bash: 1: command not found
$ eval echo \${$n} ## Third attempt to echo $1 using 'eval' succeeds
one
What exactly is happening here and how do the dollar sign and the backslash tie into the problem?
eval takes a string as its argument, and evaluates it as if you'd typed that string on a command line. (If you pass several arguments, they are first joined with spaces between them.)
${$n} is a syntax error in bash. Inside the braces, you can only have a variable name, with some possible prefix and suffixes, but you can't have arbitrary bash syntax and in particular you can't use variable expansion. There is a way of saying “the value of the variable whose name is in this variable”, though:
echo ${!n}
one
$(…) runs the command specified inside the parentheses in a subshell (i.e. in a separate process that inherits all settings such as variable values from the current shell), and gathers its output. So echo $($n) runs $n as a shell command, and displays its output. Since $n evaluates to 1, $($n) attempts to run the command 1, which does not exist.
eval echo \${$n} runs the parameters passed to eval. After expansion, the parameters are echo and ${1}. So eval echo \${$n} runs the command echo ${1}.
Note that most of the time, you must use double quotes around variable substitutions and command substitutions (i.e. anytime there's a $): "$foo", "$(foo)". Always put double quotes around variable and command substitutions, unless you know you need to leave them off. Without the double quotes, the shell performs field splitting (i.e. it splits value of the variable or the output from the command into separate words) and then treats each word as a wildcard pattern. For example:
$ ls
file1 file2 otherfile
$ set -- 'f* *'
$ echo "$1"
f* *
$ echo $1
file1 file2 file1 file2 otherfile
$ n=1
$ eval echo \${$n}
file1 file2 file1 file2 otherfile
$eval echo \"\${$n}\"
f* *
$ echo "${!n}"
f* *
eval is not used very often. In some shells, the most common use is to obtain the value of a variable whose name is not known until runtime. In bash, this is not necessary thanks to the ${!VAR} syntax. eval is still useful when you need to construct a longer command containing operators, reserved words, etc.
Simply think of eval as "evaluating your expression one additional time before execution"
eval echo \${$n} becomes echo $1 after the first round of evaluation. Three changes to notice:
The \$ became $ (The backslash is needed, otherwise it tries to evaluate ${$n}, which means a variable named {$n}, which is not allowed)
$n was evaluated to 1
The eval disappeared
In the second round, it is basically echo $1 which can be directly executed.
So eval <some command> will first evaluate <some command> (by evaluate here I mean substitute variables, replace escaped characters with the correct ones etc.), and then run the resultant expression once again.
eval is used when you want to dynamically create variables, or to read outputs from programs specifically designed to be read like this. See Eval command and security issues for examples. The link also contains some typical ways in which eval is used, and the risks associated with it.
In my experience, a "typical" use of eval is for running commands that generate shell commands to set environment variables.
Perhaps you have a system that uses a collection of environment variables, and you have a script or program that determines which ones should be set and their values. Whenever you run a script or program, it runs in a forked process, so anything it does directly to environment variables is lost when it exits. But that script or program can send the export commands to standard output.
Without eval, you would need to redirect standard output to a temporary file, source the temporary file, and then delete it. With eval, you can just:
eval "$(script-or-program)"
Note the quotes are important. Take this (contrived) example:
# activate.sh
echo 'I got activated!'
# test.py
print("export foo=bar/baz/womp")
print(". activate.sh")
$ eval $(python test.py)
bash: export: `.': not a valid identifier
bash: export: `activate.sh': not a valid identifier
$ eval "$(python test.py)"
I got activated!
The eval statement tells the shell to take eval’s arguments as commands and run them through the command-line. It is useful in a situation like below:
In your script if you are defining a command into a variable and later on you want to use that command then you should use eval:
a="ls | more"
$a
Output:
bash: command not found: ls | more
The above command didn't work as ls tried to list file with name pipe (|) and more. But these files are not there:
eval $a
Output:
file.txt
mailids
remote_cmd.sh
sample.txt
tmp
Update: Some people say one should -never- use eval. I disagree. I think the risk arises when corrupt input can be passed to eval. However there are many common situations where that is not a risk, and therefore it is worth knowing how to use eval in any case. This stackoverflow answer explains the risks of eval and alternatives to eval. Ultimately it is up to the user to determine if/when eval is safe and efficient to use.
The bash eval statement allows you to execute lines of code calculated or acquired, by your bash script.
Perhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order. That's essentially the same behavior as the bash source statement, which is what one would use, unless it was necessary to perform some kind of transformation (e.g. filtering or substitution) on the content of the imported script.
I rarely have needed eval, but I have found it useful to read or write variables whose names were contained in strings assigned to other variables. For example, to perform actions on sets of variables, while keeping the code footprint small and avoiding redundancy.
eval is conceptually simple. However, the strict syntax of the bash language, and the bash interpreter's parsing order can be nuanced and make eval appear cryptic and difficult to use or understand. Here are the essentials:
The argument passed to eval is a string expression that is calculated at runtime. eval will execute the final parsed result of its argument as an actual line of code in your script.
Syntax and parsing order are stringent. If the result isn't an executable line of bash code, in scope of your script, the program will crash on the eval statement as it tries to execute garbage.
When testing you can replace the eval statement with echo and look at what is displayed. If it is legitimate code in the current context, running it through eval will work.
The following examples may help clarify how eval works...
Example 1:
eval statement in front of 'normal' code is a NOP
$ eval a=b
$ eval echo $a
b
In the above example, the first eval statements has no purpose and can be eliminated. eval is pointless in the first line because there is no dynamic aspect to the code, i.e. it already parsed into the final lines of bash code, thus it would be identical as a normal statement of code in the bash script. The 2nd eval is pointless too, because, although there is a parsing step converting $a to its literal string equivalent, there is no indirection (e.g. no referencing via string value of an actual bash noun or bash-held script variable), so it would behave identically as a line of code without the eval prefix.
Example 2:
Perform var assignment using var names passed as string values.
$ key="mykey"
$ val="myval"
$ eval $key=$val
$ echo $mykey
myval
If you were to echo $key=$val, the output would be:
mykey=myval
That, being the final result of string parsing, is what will be executed by eval, hence the result of the echo statement at the end...
Example 3:
Adding more indirection to Example 2
$ keyA="keyB"
$ valA="valB"
$ keyB="that"
$ valB="amazing"
$ eval eval \$$keyA=\$$valA
$ echo $that
amazing
The above is a bit more complicated than the previous example, relying more heavily on the parsing-order and peculiarities of bash. The eval line would roughly get parsed internally in the following order (note the following statements are pseudocode, not real code, just to attempt to show how the statement would get broken down into steps internally to arrive at the final result).
eval eval \$$keyA=\$$valA # substitution of $keyA and $valA by interpreter
eval eval \$keyB=\$valB # convert '$' + name-strings to real vars by eval
eval $keyB=$valB # substitution of $keyB and $valB by interpreter
eval that=amazing # execute string literal 'that=amazing' by eval
If the assumed parsing order doesn't explain what eval is doing enough, the third example may describe the parsing in more detail to help clarify what is going on.
Example 4:
Discover whether vars, whose names are contained in strings, themselves contain string values.
a="User-provided"
b="Another user-provided optional value"
c=""
myvarname_a="a"
myvarname_b="b"
myvarname_c="c"
for varname in "myvarname_a" "myvarname_b" "myvarname_c"; do
eval varval=\$$varname
if [ -z "$varval" ]; then
read -p "$varname? " $varname
fi
done
In the first iteration:
varname="myvarname_a"
Bash parses the argument to eval, and eval sees literally this at runtime:
eval varval=\$$myvarname_a
The following pseudocode attempts to illustrate how bash interprets the above line of real code, to arrive at the final value executed by eval. (the following lines descriptive, not exact bash code):
1. eval varval="\$" + "$varname" # This substitution resolved in eval statement
2. .................. "$myvarname_a" # $myvarname_a previously resolved by for-loop
3. .................. "a" # ... to this value
4. eval "varval=$a" # This requires one more parsing step
5. eval varval="User-provided" # Final result of parsing (eval executes this)
Once all the parsing is done, the result is what is executed, and its effect is obvious, demonstrating there is nothing particularly mysterious about eval itself, and the complexity is in the parsing of its argument.
varval="User-provided"
The remaining code in the example above simply tests to see if the value assigned to $varval is null, and, if so, prompts the user to provide a value.
I originally intentionally never learned how to use eval, because most people will recommend to stay away from it like the plague. However I recently discovered a use case that made me facepalm for not recognizing it sooner.
If you have cron jobs that you want to run interactively to test, you might view the contents of the file with cat, and copy and paste the cron job to run it. Unfortunately, this involves touching the mouse, which is a sin in my book.
Lets say you have a cron job at /etc/cron.d/repeatme with the contents:
*/10 * * * * root program arg1 arg2
You cant execute this as a script with all the junk in front of it, but we can use cut to get rid of all the junk, wrap it in a subshell, and execute the string with eval
eval $( cut -d ' ' -f 6- /etc/cron.d/repeatme)
The cut command only prints out the 6th field of the file, delimited by spaces. Eval then executes that command.
I used a cron job here as an example, but the concept is to format text from stdout, and then evaluate that text.
The use of eval in this case is not insecure, because we know exactly what we will be evaluating before hand.
I've recently had to use eval to force multiple brace expansions to be evaluated in the order I needed. Bash does multiple brace expansions from left to right, so
xargs -I_ cat _/{11..15}/{8..5}.jpg
expands to
xargs -I_ cat _/11/8.jpg _/11/7.jpg _/11/6.jpg _/11/5.jpg _/12/8.jpg _/12/7.jpg _/12/6.jpg _/12/5.jpg _/13/8.jpg _/13/7.jpg _/13/6.jpg _/13/5.jpg _/14/8.jpg _/14/7.jpg _/14/6.jpg _/14/5.jpg _/15/8.jpg _/15/7.jpg _/15/6.jpg _/15/5.jpg
but I needed the second brace expansion done first, yielding
xargs -I_ cat _/11/8.jpg _/12/8.jpg _/13/8.jpg _/14/8.jpg _/15/8.jpg _/11/7.jpg _/12/7.jpg _/13/7.jpg _/14/7.jpg _/15/7.jpg _/11/6.jpg _/12/6.jpg _/13/6.jpg _/14/6.jpg _/15/6.jpg _/11/5.jpg _/12/5.jpg _/13/5.jpg _/14/5.jpg _/15/5.jpg
The best I could come up with to do that was
xargs -I_ cat $(eval echo _/'{11..15}'/{8..5}.jpg)
This works because the single quotes protect the first set of braces from expansion during the parsing of the eval command line, leaving them to be expanded by the subshell invoked by eval.
There may be some cunning scheme involving nested brace expansions that allows this to happen in one step, but if there is I'm too old and stupid to see it.
You asked about typical uses.
One common complaint about shell scripting is that you (allegedly) can't pass by reference to get values back out of functions.
But actually, via "eval", you can pass by reference. The callee can pass back a list of variable assignments to be evaluated by the caller. It is pass by reference because the caller can allowed to specify the name(s) of the result variable(s) - see example below. Error results can be passed back standard names like errno and errstr.
Here is an example of passing by reference in bash:
#!/bin/bash
isint()
{
re='^[-]?[0-9]+$'
[[ $1 =~ $re ]]
}
#args 1: name of result variable, 2: first addend, 3: second addend
iadd()
{
if isint ${2} && isint ${3} ; then
echo "$1=$((${2}+${3}));errno=0"
return 0
else
echo "errstr=\"Error: non-integer argument to iadd $*\" ; errno=329"
return 1
fi
}
var=1
echo "[1] var=$var"
eval $(iadd var A B)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[2] var=$var (unchanged after error)"
eval $(iadd var $var 1)
if [[ $errno -ne 0 ]]; then
echo "errstr=$errstr"
echo "errno=$errno"
fi
echo "[3] var=$var (successfully changed)"
The output looks like this:
[1] var=1
errstr=Error: non-integer argument to iadd var A B
errno=329
[2] var=1 (unchanged after error)
[3] var=2 (successfully changed)
There is almost unlimited band width in that text output! And there are more possibilities if the multiple output lines are used: e.g., the first line could be used for variable assignments, the second for continuous 'stream of thought', but that's beyond the scope of this post.
In the question:
who | grep $(tty | sed s:/dev/::)
outputs errors claiming that files a and tty do not exist. I understood this to mean that tty is not being interpreted before execution of grep, but instead that bash passed tty as a parameter to grep, which interpreted it as a file name.
There is also a situation of nested redirection, which should be handled by matched parentheses which should specify a child process, but bash is primitively a word separator, creating parameters to be sent to a program, therefore parentheses are not matched first, but interpreted as seen.
I got specific with grep, and specified the file as a parameter instead of using a pipe. I also simplified the base command, passing output from a command as a file, so that i/o piping would not be nested:
grep $(tty | sed s:/dev/::) <(who)
works well.
who | grep $(echo pts/3)
is not really desired, but eliminates the nested pipe and also works well.
In conclusion, bash does not seem to like nested pipping. It is important to understand that bash is not a new-wave program written in a recursive manner. Instead, bash is an old 1,2,3 program, which has been appended with features. For purposes of assuring backward compatibility, the initial manner of interpretation has never been modified. If bash was rewritten to first match parentheses, how many bugs would be introduced into how many bash programs? Many programmers love to be cryptic.
As clearlight has said, "(p)erhaps the most straightforward example would be a bash program that opens another bash script as a text file, reads each line of text, and uses eval to execute them in order". I'm no expert, but the textbook I'm currently reading (Shell-Programmierung by Jürgen Wolf) points to one particular use of this that I think would be a valuable addition to the set of potential use cases collected here.
For debugging purposes, you may want to go through your script line by line (pressing Enter for each step). You could use eval to execute every line by trapping the DEBUG signal (which I think is sent after every line):
trap 'printf "$LINENO :-> " ; read line ; eval $line' DEBUG
I like the "evaluating your expression one additional time before execution" answer, and would like to clarify with another example.
var="\"par1 par2\""
echo $var # prints nicely "par1 par2"
function cntpars() {
echo " > Count: $#"
echo " > Pars : $*"
echo " > par1 : $1"
echo " > par2 : $2"
if [[ $# = 1 && $1 = "par1 par2" ]]; then
echo " > PASS"
else
echo " > FAIL"
return 1
fi
}
# Option 1: Will Pass
echo "eval \"cntpars \$var\""
eval "cntpars $var"
# Option 2: Will Fail, with curious results
echo "cntpars \$var"
cntpars $var
The curious results in option 2 are that we would have passed two parameters as follows:
First parameter: "par1
Second parameter: par2"
How is that for counter intuitive? The additional eval will fix that.
It was adapted from another answer on How can I reference a file for variables using Bash?

Why won't my grep-filtered string print from within a while-loop?

Tried to keep my code as simple as possible:
1: What are the rules for using echo within a while loop?
All my $a and some of my $word variables are echoed not my echo kk?
2: What is the scope of my count variable? Why is it not working within my while loop? can I extend the variable to make it global?
3: When I use the grep in the final row the $word cariable only prints the first word in the passing rows ehile if I remove the grep line in the end $work functions as intended and prints all the words.
count=1
while read a; do
((count=count+1))
if [ $count -le 2 ]
then
echo $a
echo kk
for word in $a; do
echo $word
done
fi
done < data.txt | grep Iteration
Use Process Substitution
In a comment, you say:
I thtought I was using grep on data.txt (sic)
No. Your current pipeline passes the loop's results through grep, not the source file. To do that, you need to rewrite your redirection to use process substitution. For example:
count=1
while read a; do
((count=count+1))
if [ $count -le 2 ]
then
echo $a
echo kk
for word in $a; do
echo $word
done
fi
done < <(fgrep Iteration data.txt)
#CodeGnome answered your question but there's other problems with your script that will come back to bite you at some point. (see https://unix.stackexchange.com/questions/169716/why-is-using-a-shell-loop-to-process-text-considered-bad-practice for discussions on some of them and also google quoting shell variables). Just don't do it. Shell scripts are just for sequencing calls to tools and the UNIX tool for manipulating text is awk. In this case all you'd need to do the job robustly, portably and efficiently would be:
awk '
/Iteration/ {
if (++count <= 2) {
print
print "kk"
for (i=1; i<=NF; i++) {
print $i
}
}
}' data.txt
and of course it'd be more efficient still if you just stop reading the input when count hits 2:
awk '
/Iteration/ {
print
print "kk"
for (i=1; i<=NF; i++) {
print $i
}
if (++count == 2) {
exit
}
}' data.txt
To complement CodeGnome's helpful answer with an explanation of how your command actually works and why it doesn't do what you want:
In Bash's grammar, an input redirection such as < data.txt is part of a single command, whereas |, the pipe symbol, chains multiple commands, from left to right, to form a pipeline.
Technically, while ... done ... < data.txt | grep Iteration is a single pipeline composed of 2 commands:
a single compound command (while ...; do ...; done) with an input redirection (< data.txt),
and a simple command (grep Iteration) that receives the stdout output from the compound command via its stdin, courtesy of the pipe.
In other words:
only the contents of data.txt is fed to the while loop as input (via stdin),
and whatever stdout output the while loop produces is then sent to the next pipeline segment, the grep command.
By contrast, it sounds like you want to apply grep to data.txt first, and only sent the matching lines to the while loop.
You have the following options for sending a command's output to another command:
Note: The following solutions use a simplified while loop for brevity - whether a while command is single-line or spans multiple lines is irrelevant.
Also, instead of using input redirection (< data.txt) to pass the file content to grep, data.txt is passed as a filename argument.
Option 1: Place the command whose output to send to your while loop first in the pipeline:
grep 'Iteration' data.txt | while read -r a; do echo "$a"; done
The down-side of this approach is that your while loop then runs in a subshell (as all segments of a pipeline do by default), which means that variables defined or modified in your while command won't be visible to the current shell.
In Bash v4.2+, you can fix this by running shopt -s lastpipe, which tells Bash to run the last pipeline segment - the while command in this case - in the current shell instead.
Note that lastpipe is a nonstandard bash extension to the POSIX standard.
(To try this in an interactive shell, you must first turn off job control with set +m.)
Option 2: Use a process substitution:
Loosely speaking, a process substitution <(...) allows you to present command output as the content of a temporary file that cleans up after itself.
Since <(...) expands to the temporary file's (FIFO's) path, and read in the while loop only accepts stdin input, input redirection must be applied as well: < <(...):
while read -r a; do echo "$a"; done < <(grep 'Iteration' data.txt)
The advantage of this approach is that the while loop runs in the current subshell, and any variables definitions or modifications therefore remain in scope after the command completes.
The potential down-side of this approach is that process substitutions are a nonstandard bash extension to the POSIX standard (although ksh and zsh support them too).
Option 3: Use a command substitution inside a here-document:
Using the command first in the pipeline (option 1) is a POSIX-compliant approach, but doesn't allow you to modify variables in the current shell (and Bash's lastpipe option is not POSIX-compliant).
The only POSIX-compliant way to send command output to a command that runs in the current shell is to use a command substitution ($(...)) inside a double-quoted here-document:
while read -r a; do echo "$a"; done <<EOF
$(grep 'Iteration' data.txt)
EOF
Streamlining your code and making it more robust:
The rest of your code has some non-obvious pitfalls that are worth addressing:
Double-quote your variable references (e.g., echo "$a" instead of echo $a), unless you specifically want word-splitting and globbing (filename expansion) applied to the values; word splitting and globbing are two kinds of shell expansions.
Similarly, don't use for to iterate over an (of necessity unquoted) variable reference (don't use for word in $a, in your case), unless you want globbing applied to the individual words - see what happens when you run $a='one *'; for word in $a; do echo "$word"; done
You could turn globbing off beforehand (set -f) and back on after (set +f), but it's better to use read -ra words ... to read the words into an array first, and then safely iterate over the array elements with for word in "${words[#]}"; ...- note the "..." around the array variable reference.
Always use -r with read; without it, rarely used \-preprocessing is applied, which will "eat" embedded \ chars.
If we heed the advice above, apply a few additional tweaks, and use a process substitution to feed grep's output to the while loop, we get:
count=1
while read -r a; do # Note the -r
if (( ++count <= 2 )); then
echo "$a"
# Split $a safely into words and store the words in
# array variable ${words[#]}.
read -ra words <<<"$a" # Note the -a to read into an *array*.
# Loop over the words (elements of the array).
# Note: To simply print the words, you could use
# `printf '%s\n' "${words[#]}"`` instead of the loop.
for word in "${words[#]}"; do
echo "$word"
done
fi
done < <(grep 'Iteration' data.txt)
Note: As written, you don't need a loop at all, because you always exit after the 1st iteration.
Finally, as a general alternative for larger input sets, consider Ed Morton's helpful answer, which is much faster due to using awk to process your input file, whereas looping in shell code is generally slow.

String expansion - escaped quoted variable to value

To get started, here's the script I'm running to get the offending string:
# sed finds all sourced file paths from inputted file.
#
# while reads each match output from sed to $SOURCEFILE variable.
# Each should be a file path, or a variable that represents a file path.
# Any variables found should be expanded to the full path.
#
# echo and calls are used for demonstractive purposes only
# I intend to do something else with the path once it's expanded.
PATH_SOME_SCRIPT="/path/to/bash/script"
while read -r SOURCEFILE; do
echo "$SOURCEFILE"
"$SOURCEFILE"
$SOURCEFILE
done < <(cat $PATH_SOME_SCRIPT | sed -n -e "s/^\(source\|\.\|\$include\) //p")
You may also wish to use the following to test this out as mock data:
[ /path/to/bash/script ]
#!/bin/bash
source "$HOME/bash_file"
source "$GLOBAL_VAR_SCRIPT_PATH"
echo "No cow powers here"
For the tl;dr crew, basically the while loop spits out the following on the mock data:
"$HOME/bash_file"
bash: "$HOME/bash_file": no such file or directory
bash: "$HOME/bash_file": no such file or directory
"$GLOBAL_VAR_SCRIPT_PATH"
"$GLOBAL_VAR_SCRIPT_PATH": command not found
"$GLOBAL_VAR_SCRIPT_PATH": command not found
My question is, can you get the variable to expand correctly, e.g., print "/home//bash_file" and "/expanded/variable/path"? I should also state that although eval works I do not intend to use it because of its potential insecurities.
Protip that any variable value used in cat | sed would be available globally, including to the calling script, so it's not because the script cannot call the variable value.
FIRST SOLUTION ATTEMPT
Using anubhava's envsubst solution:
SOMEVARIABLE="/home/nick/.some_path"
while read -r SOURCEFILE; do
echo "$SOURCEFILE"
envsubst <<< "$SOURCEFILE";
done < <(echo -e "\"\$SOMEVARIABLE\"\n\"$HOME/.another_file\"")
This outputs the following:
"$SOMEVARIABLE"
""
"/home/nick/.another_file"
"/home/nick/.another_file"
Unfortunately, it does not expand the variable! Oh dear :(
SECOND SOLUTION ATTEMPT
Based upon the first attempt:
export SOMEVARIABLE="/home/nick/.some_path"
while read -r SOURCEFILE; do
echo "$SOURCEFILE"
envsubst <<< "$SOURCEFILE";
done < <(echo -e "\"\$SOMEVARIABLE\"\n\"$HOME/.another_file\"")
unset SOMEVARIABLE
which produces the results we wanted without eval and without messing with global variables (for too long anyway), hoorah!
Good runner-ups were further suggested using eval (although potentially unsafe) which can be found in this answer and here (link courtesy of anubhava's extended comments).
My question is, can you get the variable to expand correctly, e.g., print "/home//bash_file" and "/expanded/variable/path"?
Yes you can use envsubst program, that substitutes the values of environment variables:
while read -r sourceFile; do
envsubst <<< "$sourceFile"
done < <(sed -n "s/^\(source\|\.\|\$include\) //p" "$PATH_SOME_SCRIPT")
I think you are asking how to recursively expand variables in bash. Try
expanded=$(eval echo $SOURCEFILE)
inside your loop. eval runs the expanded command you give it. Since $SOURCEFILE isn't in quotes, it will be expanded to, e.g., $HOME/whatever. Then the eval will expand the $HOME before passing it to echo. echo will print the result, and expanded=$(...) will put the printed result in $expanded.

shell variable not keeping value [duplicate]

This question already has answers here:
Variables getting reset after the while read loop that reads from a pipeline
(3 answers)
Closed 7 years ago.
I have 3 lines in file /root/backuplist.txt.
The first echo prints perfectly, but the last one prints an empty line; I'm not sure why. Somehow, the $DIRS value is getting unset.
#!/bin/bash
cat /root/backuplist.txt |
while read line
do
DIRS+="$line "
echo $DIRS
done
echo $DIRS
Problem is use of pipe here, which is forking a sub-shell for your while loop and thus changes to DIRS are being made in the child shell that are not visible in the parent shell. Besides cat is unnecessary here.
Have it this way:
#!/bin/bash
while read -r line
do
DIRS+="$line "
echo "$DIRS"
done < /root/backuplist.txt
echo "$DIRS"

How to use the read command in Bash?

When I try to use the read command in Bash like this:
echo hello | read str
echo $str
Nothing echoed, while I think str should contain the string hello. Can anybody please help me understand this behavior?
The read in your script command is fine. However, you execute it in the pipeline, which means it is in a subshell, therefore, the variables it reads to are not visible in the parent shell. You can either
move the rest of the script in the subshell, too:
echo hello | { read str
echo $str
}
or use command substitution to get the value of the variable out of the subshell
str=$(echo hello)
echo $str
or a slightly more complicated example (Grabbing the 2nd element of ls)
str=$(ls | { read a; read a; echo $a; })
echo $str
Other bash alternatives that do not involve a subshell:
read str <<END # here-doc
hello
END
read str <<< "hello" # here-string
read str < <(echo hello) # process substitution
Typical usage might look like:
i=0
echo -e "hello1\nhello2\nhello3" | while read str ; do
echo "$((++i)): $str"
done
and output
1: hello1
2: hello2
3: hello3
The value disappears since the read command is run in a separate subshell: Bash FAQ 24
To put my two cents here: on KSH, reading as is to a variable will work, because according to the IBM AIX documentation, KSH's read does affects the current shell environment:
The setting of shell variables by the read command affects the current shell execution environment.
This just resulted in me spending a good few minutes figuring out why a one-liner ending with read that I've used a zillion times before on AIX didn't work on Linux... it's because KSH does saves to the current environment and BASH doesn't!
I really only use read with "while" and a do loop:
echo "This is NOT a test." | while read -r a b c theRest; do
echo "$a" "$b" "$theRest"; done
This is a test.
For what it's worth, I have seen the recommendation to always use -r with the read command in bash.
You don't need echo to use read
read -p "Guess a Number" NUMBER
Another alternative altogether is to use the printf function.
printf -v str 'hello'
Moreover, this construct, combined with the use of single quotes where appropriate, helps to avoid the multi-escape problems of subshells and other forms of interpolative quoting.
Do you need the pipe?
echo -ne "$MENU"
read NUMBER

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