How to expand variables using previously defined variables inside a file - bash

I have a property file having multiple key value pairs. Some of the values uses previously defined keys. Following is the sample
xpath=abc/temp.txt
fullpath=$HOME/$xpath
...
I want to parse this file line by line and print the lines along with resolving env variables like $HOME as well as earlier defined variables like $xpath
The expected output is
xpath=abc/temp.txt
fullpath=tempuser/abc/temp.txt
...
How do I expand the variables in this way in a bash script

If you want to parse line by line and interpret the variables, You maybe want to use the eval for evaluating the String, like:
while IFS='=' read -r key value
do
key=$(echo $key | tr '.' '_')
if [[ ! -z $key ]]
then
v=`eval "echo ${value}"`
eval "${key}='${v}'"
echo "${key}=${v}"
fi
done < "my.properties"
In the above code snippet, use the eval with echo to interpret the variables.

Since the assignments are valid bash, source settings.env suffices to evaluate them. However, you also want to print the assignments, so more trickery required:
PS4="\000" source <(echo 'set -x'; cat settings.env; echo '{ set +x; } 2>/dev/null')
This trick uses the bash debugging facility of set -x to print the output of each assignment as it's performed. In words:
PS4="\000" removes the debugging prompt, which by default is +
<() creates a new file, by the pipeline contained within the parentheses
set -x enables debugging
cat settings.env inserts your settings
{ set +x; } 2>dev/null disables debugging, without outputting it doing so

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?

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.

How to pass shell variables in "echo" command

I have she script as the below content
chr=$0
start=$1
end=$2
echo -e "$chr\t$start\t$end" > covdb_input.bed
How do i pass the chr,Start and end variables in to echo command.. or write same to file "covdb_input.bed" with TAB sep as in echo command.
You're doing everything right, except that you probably initialize your variables with the wrong things.
I'm assuming you get arguments for the script (or shell function), and that you want to use these. Then pick the positional variables from $1 and onwards as $0 will usually contain the name of the current shell script or shell function.
Also, you might find people scoffing about the use of -e with echo (it's a common but non-standard option). Instead of using echo you could use printf like this:
printf "%s\t%s\t%s" "$chr" "$start" "$end" >myfile.bed
Or just
printf "$chr\t$start\t$end" >myfile.bed

bash - cleaner way to count the number of lines in a variable using pure bash?

I have a variable containing some command output, and I want to count the number of lines in that output. I am trying to do this using pure bash (instead of piping to wc like in this question).
The best I have come up with seems kind of cumbersome:
function count_lines() {
num_lines=0
while IFS='' read -r line; do
((num_lines++))
done <<< "$1"
echo "$num_lines"
}
count_lines "$my_var"
Is there a cleaner or shorter way to do this?
count_lines () (
IFS=$'\n'
set -f
set -- $1
echo $#
)
Some tricks used:
The function is defined using a subshell instead of a command group to localize the change to IFS and the -f shell option. (You could be less fancy and use local IFS=$'\n' instead, and running set +f at the end of the function).
Disable filename generation to avoid any metacharacters in the argument from expanding and interfering with the line count.
Once IFS is changed, set the positional parameters for the function using unquoted parameter expansion for the argument.
Finally, output the number of positional parameters; there is one per line in the original argument.
IFS=$'\n'
set -f
x=( $variable )
Now ${#x[#]} will contain the number of 'lines'. (Use "set +f" to undo "set -f".)
Another possible solution I came up with:
export x=0;while read i;do export x=$(($x+1));done < /path/to/your/file;echo $x

How to read stdin when no arguments are passed?

Script doesn't work when I want to use standard input when there are no arguments (files) passed. Is there any way how to use stdin instead of a file in this code?
I tried this:
if [ ! -n $1 ] # check if argument exists
then
$1=$(</dev/stdin) # if not use stdin as an argument
fi
var="$1"
while read line
do
... # find the longest line
done <"$var"
For a general case of wanting to read a value from stdin when a parameter is missing, this will work.
$ echo param | script.sh
$ script.sh param
script.sh
#!/bin/bash
set -- "${1:-$(</dev/stdin)}" "${#:2}"
echo $1
Just substitute bash's specially interpreted /dev/stdin as the filename:
VAR=$1
while read blah; do
...
done < "${VAR:-/dev/stdin}"
(Note that bash will actually use that special file /dev/stdin if built for an OS that offers it, but since bash 2.04 will work around that file's absence on systems that do not support it.)
pilcrow's answer provides an elegant solution; this is an explanation of why the OP's approach didn't work.
The main problem with the OP's approach was the attempt to assign to positional parameter $1 with $1=..., which won't work.
The LHS is expanded by the shell to the value of $1, and the result is interpreted as the name of the variable to assign to - clearly, not the intent.
The only way to assign to $1 in bash is via the set builtin.
The caveat is that set invariably sets all positional parameters, so you have to include the other ones as well, if any.
set -- "${1:-/dev/stdin}" "${#:2}" # "${#:2}" expands to all remaining parameters
(If you expect only at most 1 argument, set -- "${1:-/dev/stdin}" will do.)
The above also corrects a secondary problem with the OP's approach: the attempt to store the contents rather than the filename of stdin in $1, since < is used.
${1:-/dev/stdin} is an application of bash parameter expansion that says: return the value of $1, unless $1 is undefined (no argument was passed) or its value is the empty string (""or '' was passed). The variation ${1-/dev/stdin} (no :) would only return /dev/stdin if $1 is undefined (if it contains any value, even the empty string, it would be returned).
If we put it all together:
# Default to filename '/dev/stdin' (stdin), if none was specified.
set -- "${1:-/dev/stdin}" "${#:2}"
while read -r line; do
... # find the longest line
done < "$1"
But, of course, the much simpler approach would be to use ${1:-/dev/stdin} as the filename directly:
while read -r line; do
... # find the longest line
done < "${1:-/dev/stdin}"
or, via an intermediate variable:
filename=${1:-/dev/stdin}
while read -r line; do
... # find the longest line
done < "$filename"
Variables are assigned a value by Var=Value and that variable is used by e.g. echo $Var. In your case, that would amount to
1=$(</dev/stdin)
when assigning the standard input. However, I do not think that variable names are allowed to start with a digit character. See the question bash read from file or stdin for ways to solve this.
Here is my version of script:
#!/bin/bash
file=${1--} # POSIX-compliant; ${1:--} can be used either.
while IFS= read -r line; do
printf '%s\n' "$line"
done < <(cat -- "$file")
If file is not present in the argument, read the from standard input.
See more examples: How to read from file or stdin in bash? at stackoverflow SE

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