Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
Related
Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
In the following program, if I set the variable $foo to the value 1 inside the first if statement, it works in the sense that its value is remembered after the if statement. However, when I set the same variable to the value 2 inside an if which is inside a while statement, it's forgotten after the while loop. It's behaving like I'm using some sort of copy of the variable $foo inside the while loop and I am modifying only that particular copy. Here's a complete test program:
#!/bin/bash
set -e
set -u
foo=0
bar="hello"
if [[ "$bar" == "hello" ]]
then
foo=1
echo "Setting \$foo to 1: $foo"
fi
echo "Variable \$foo after if statement: $foo"
lines="first line\nsecond line\nthird line"
echo -e $lines | while read line
do
if [[ "$line" == "second line" ]]
then
foo=2
echo "Variable \$foo updated to $foo inside if inside while loop"
fi
echo "Value of \$foo in while loop body: $foo"
done
echo "Variable \$foo after while loop: $foo"
# Output:
# $ ./testbash.sh
# Setting $foo to 1: 1
# Variable $foo after if statement: 1
# Value of $foo in while loop body: 1
# Variable $foo updated to 2 inside if inside while loop
# Value of $foo in while loop body: 2
# Value of $foo in while loop body: 2
# Variable $foo after while loop: 1
# bash --version
# GNU bash, version 4.1.10(4)-release (i686-pc-cygwin)
echo -e $lines | while read line
...
done
The while loop is executed in a subshell. So any changes you do to the variable will not be available once the subshell exits.
Instead you can use a here string to re-write the while loop to be in the main shell process; only echo -e $lines will run in a subshell:
while read line
do
if [[ "$line" == "second line" ]]
then
foo=2
echo "Variable \$foo updated to $foo inside if inside while loop"
fi
echo "Value of \$foo in while loop body: $foo"
done <<< "$(echo -e "$lines")"
You can get rid of the rather ugly echo in the here-string above by expanding the backslash sequences immediately when assigning lines. The $'...' form of quoting can be used there:
lines=$'first line\nsecond line\nthird line'
while read line; do
...
done <<< "$lines"
UPDATED#2
Explanation is in Blue Moons's answer.
Alternative solutions:
Eliminate echo
while read line; do
...
done <<EOT
first line
second line
third line
EOT
Add the echo inside the here-is-the-document
while read line; do
...
done <<EOT
$(echo -e $lines)
EOT
Run echo in background:
coproc echo -e $lines
while read -u ${COPROC[0]} line; do
...
done
Redirect to a file handle explicitly (Mind the space in < <!):
exec 3< <(echo -e $lines)
while read -u 3 line; do
...
done
Or just redirect to the stdin:
while read line; do
...
done < <(echo -e $lines)
And one for chepner (eliminating echo):
arr=("first line" "second line" "third line");
for((i=0;i<${#arr[*]};++i)) { line=${arr[i]};
...
}
Variable $lines can be converted to an array without starting a new sub-shell. The characters \ and n has to be converted to some character (e.g. a real new line character) and use the IFS (Internal Field Separator) variable to split the string into array elements. This can be done like:
lines="first line\nsecond line\nthird line"
echo "$lines"
OIFS="$IFS"
IFS=$'\n' arr=(${lines//\\n/$'\n'}) # Conversion
IFS="$OIFS"
echo "${arr[#]}", Length: ${#arr[*]}
set|grep ^arr
Result is
first line\nsecond line\nthird line
first line second line third line, Length: 3
arr=([0]="first line" [1]="second line" [2]="third line")
You are asking this bash FAQ. The answer also describes the general case of variables set in subshells created by pipes:
E4) If I pipe the output of a command into read variable, why
doesn't the output show up in $variable when the read command finishes?
This has to do with the parent-child relationship between Unix
processes. It affects all commands run in pipelines, not just
simple calls to read. For example, piping a command's output
into a while loop that repeatedly calls read will result in
the same behavior.
Each element of a pipeline, even a builtin or shell function,
runs in a separate process, a child of the shell running the
pipeline. A subprocess cannot affect its parent's environment.
When the read command sets the variable to the input, that
variable is set only in the subshell, not the parent shell. When
the subshell exits, the value of the variable is lost.
Many pipelines that end with read variable can be converted
into command substitutions, which will capture the output of
a specified command. The output can then be assigned to a
variable:
grep ^gnu /usr/lib/news/active | wc -l | read ngroup
can be converted into
ngroup=$(grep ^gnu /usr/lib/news/active | wc -l)
This does not, unfortunately, work to split the text among
multiple variables, as read does when given multiple variable
arguments. If you need to do this, you can either use the
command substitution above to read the output into a variable
and chop up the variable using the bash pattern removal
expansion operators or use some variant of the following
approach.
Say /usr/local/bin/ipaddr is the following shell script:
#! /bin/sh
host `hostname` | awk '/address/ {print $NF}'
Instead of using
/usr/local/bin/ipaddr | read A B C D
to break the local machine's IP address into separate octets, use
OIFS="$IFS"
IFS=.
set -- $(/usr/local/bin/ipaddr)
IFS="$OIFS"
A="$1" B="$2" C="$3" D="$4"
Beware, however, that this will change the shell's positional
parameters. If you need them, you should save them before doing
this.
This is the general approach -- in most cases you will not need to
set $IFS to a different value.
Some other user-supplied alternatives include:
read A B C D << HERE
$(IFS=.; echo $(/usr/local/bin/ipaddr))
HERE
and, where process substitution is available,
read A B C D < <(IFS=.; echo $(/usr/local/bin/ipaddr))
Hmmm... I would almost swear that this worked for the original Bourne shell, but don't have access to a running copy just now to check.
There is, however, a very trivial workaround to the problem.
Change the first line of the script from:
#!/bin/bash
to
#!/bin/ksh
Et voila! A read at the end of a pipeline works just fine, assuming you have the Korn shell installed.
This is an interesting question and touches on a very basic concept in Bourne shell and subshell. Here I provide a solution that is different from the previous solutions by doing some kind of filtering. I will give an example that may be useful in real life. This is a fragment for checking that downloaded files conform to a known checksum. The checksum file look like the following (Showing just 3 lines):
49174 36326 dna_align_feature.txt.gz
54757 1 dna.txt.gz
55409 9971 exon_transcript.txt.gz
The shell script:
#!/bin/sh
.....
failcnt=0 # this variable is only valid in the parent shell
#variable xx captures all the outputs from the while loop
xx=$(cat ${checkfile} | while read -r line; do
num1=$(echo $line | awk '{print $1}')
num2=$(echo $line | awk '{print $2}')
fname=$(echo $line | awk '{print $3}')
if [ -f "$fname" ]; then
res=$(sum $fname)
filegood=$(sum $fname | awk -v na=$num1 -v nb=$num2 -v fn=$fname '{ if (na == $1 && nb == $2) { print "TRUE"; } else { print "FALSE"; }}')
if [ "$filegood" = "FALSE" ]; then
failcnt=$(expr $failcnt + 1) # only in subshell
echo "$fname BAD $failcnt"
fi
fi
done | tail -1) # I am only interested in the final result
# you can capture a whole bunch of texts and do further filtering
failcnt=${xx#* BAD } # I am only interested in the number
# this variable is in the parent shell
echo failcnt $failcnt
if [ $failcnt -gt 0 ]; then
echo $failcnt files failed
else
echo download successful
fi
The parent and subshell communicate through the echo command. You can pick some easy to parse text for the parent shell. This method does not break your normal way of thinking, just that you have to do some post processing. You can use grep, sed, awk, and more for doing so.
I use stderr to store within a loop, and read from it outside.
Here var i is initially set and read inside the loop as 1.
# reading lines of content from 2 files concatenated
# inside loop: write value of var i to stderr (before iteration)
# outside: read var i from stderr, has last iterative value
f=/tmp/file1
g=/tmp/file2
i=1
cat $f $g | \
while read -r s;
do
echo $s > /dev/null; # some work
echo $i > 2
let i++
done;
read -r i < 2
echo $i
Or use the heredoc method to reduce the amount of code in a subshell.
Note the iterative i value can be read outside the while loop.
i=1
while read -r s;
do
echo $s > /dev/null
let i++
done <<EOT
$(cat $f $g)
EOT
let i--
echo $i
How about a very simple method
+call your while loop in a function
- set your value inside (nonsense, but shows the example)
- return your value inside
+capture your value outside
+set outside
+display outside
#!/bin/bash
# set -e
# set -u
# No idea why you need this, not using here
foo=0
bar="hello"
if [[ "$bar" == "hello" ]]
then
foo=1
echo "Setting \$foo to $foo"
fi
echo "Variable \$foo after if statement: $foo"
lines="first line\nsecond line\nthird line"
function my_while_loop
{
echo -e $lines | while read line
do
if [[ "$line" == "second line" ]]
then
foo=2; return 2;
echo "Variable \$foo updated to $foo inside if inside while loop"
fi
echo -e $lines | while read line
do
if [[ "$line" == "second line" ]]
then
foo=2;
echo "Variable \$foo updated to $foo inside if inside while loop"
return 2;
fi
# Code below won't be executed since we returned from function in 'if' statement
# We aready reported the $foo var beint set to 2 anyway
echo "Value of \$foo in while loop body: $foo"
done
}
my_while_loop; foo="$?"
echo "Variable \$foo after while loop: $foo"
Output:
Setting $foo 1
Variable $foo after if statement: 1
Value of $foo in while loop body: 1
Variable $foo after while loop: 2
bash --version
GNU bash, version 3.2.51(1)-release (x86_64-apple-darwin13)
Copyright (C) 2007 Free Software Foundation, Inc.
Though this is an old question and asked several times, here's what I'm doing after hours fidgeting with here strings, and the only option that worked for me is to store the value in a file during while loop sub-shells and then retrieve it. Simple.
Use echo statement to store and cat statement to retrieve. And the bash user must chown the directory or have read-write chmod access.
#write to file
echo "1" > foo.txt
while condition; do
if (condition); then
#write again to file
echo "2" > foo.txt
fi
done
#read from file
echo "Value of \$foo in while loop body: $(cat foo.txt)"
I have a code like that
var="before"
echo "$someString" | sed '$someRegex' | while read line
do
if [ $condition ]; then
var="after"
echo "$var" #first echo
fi
done
echo "$var" #second echo
Here first echo print "after", but second is "before". How can I make second echo print "after". I think it is because of pipe buy I don't know how figure out.
Thanks for any solutions...
answer edit:
I corrected it and it works fine. Thanks eugene for your useful answer
var="before"
while read line
do
if [ $condition ]; then
var="after"
echo "$var" #first echo
fi
done < <(echo "$someString" | sed '$someRegex')
echo "$var" #second echo
The reason for this behaviour is that a while loop runs in a subshell when it's part of a pipeline. For the while loop above, a new subshell with its own copy of the variable var is created.
See this article for possible workarounds: I set variables in a loop that's in a pipeline. Why do they disappear after the loop terminates? Or, why can't I pipe data to read?.
Bash allows to use: cat <(echo "$FILECONTENT")
Bash also allow to use: while read i; do echo $i; done </etc/passwd
to combine previous two this can be used: echo $FILECONTENT | while read i; do echo $i; done
The problem with last one is that it creates sub-shell and after the while loop ends variable i cannot be accessed any more.
My question is:
How to achieve something like this: while read i; do echo $i; done <(echo "$FILECONTENT") or in other words: How can I be sure that i survives while loop?
Please note that I am aware of enclosing while statement into {} but this does not solves the problem (imagine that you want use the while loop in function and return i variable)
The correct notation for Process Substitution is:
while read i; do echo $i; done < <(echo "$FILECONTENT")
The last value of i assigned in the loop is then available when the loop terminates.
An alternative is:
echo $FILECONTENT |
{
while read i; do echo $i; done
...do other things using $i here...
}
The braces are an I/O grouping operation and do not themselves create a subshell. In this context, they are part of a pipeline and are therefore run as a subshell, but it is because of the |, not the { ... }. You mention this in the question. AFAIK, you can do a return from within these inside a function.
Bash also provides the shopt builtin and one of its many options is:
lastpipe
If set, and job control is not active, the shell runs the last command of a pipeline not executed in the background in the current shell environment.
Thus, using something like this in a script makes the modfied sum available after the loop:
FILECONTENT="12 Name
13 Number
14 Information"
shopt -s lastpipe # Comment this out to see the alternative behaviour
sum=0
echo "$FILECONTENT" |
while read number name; do ((sum+=$number)); done
echo $sum
Doing this at the command line usually runs foul of 'job control is not active' (that is, at the command line, job control is active). Testing this without using a script failed.
Also, as noted by Gareth Rees in his answer, you can sometimes use a here string:
while read i; do echo $i; done <<< "$FILECONTENT"
This doesn't require shopt; you may be able to save a process using it.
Jonathan Leffler explains how to do what you want using process substitution, but another possibility is to use a here string:
while read i; do echo "$i"; done <<<"$FILECONTENT"
This saves a process.
This function makes duplicates $NUM times of jpg files (bash)
function makeDups() {
NUM=$1
echo "Making $1 duplicates for $(ls -1 *.jpg|wc -l) files"
ls -1 *.jpg|sort|while read f
do
COUNT=0
while [ "$COUNT" -le "$NUM" ]
do
cp $f ${f//sm/${COUNT}sm}
((COUNT++))
done
done
}
How do I set a variable in the parent shell, from a subshell?
a=3
(a=4)
echo $a
The whole point of a subshell is that it doesn't affect the calling session. In bash a subshell is a child process, other shells differ but even then a variable setting in a subshell does not affect the caller. By definition.
Do you need a subshell? If you just need a group then use braces:
a=3
{ a=4;}
echo $a
gives 4 (be careful of the spaces in that one). Alternatively, write the variable value to stdout and capture it in the caller:
a=3
a=$(a=4;echo $a)
echo $a
avoid using back-ticks ``, they are deprecated and can be difficult to read.
There is the gdb-bash-variable hack:
gdb --batch-silent -ex "attach $$" -ex 'set bind_variable("a", "4", 0)';
although that always sets a variable in the global scope, not just the parent scope
You don't. The subshell doesn't have access to its parent's environment. (At least within the abstraction that Bash provides. You could potentially try to use gdb, or smash the stack, or whatnot, to gain such access clandestinely. I wouldn't recommend that, though.)
One alternative is for the subshell to write assignment statements to a temporary file for its parent to read:
a=3
(echo 'a=4' > tmp)
. tmp
rm tmp
echo "$a"
If the problem is related to a while loop, one way to fix this is by using Process Substitution:
var=0
while read i;
do
# perform computations on $i
((var++))
done < <(find . -type f -name "*.bin" -maxdepth 1)
as shown here: https://stackoverflow.com/a/13727116/2547445
To change variables in a script called from a parent script, you can call the script preceded with a "."
(EDIT - for explanation)
In most shells "." is an alias for "source". the source command just inserts the text of another file at that position in the executing script. In the context of this question this answer avoids a sub-shell
a=3
echo $a
. ./calledScript.sh
echo $a
in calledScript.sh
a=4
Expected output
3
4
By reading the answer from #ruakh (thank you) with a temporary file approach and the comments asking for a file descriptors solution, I got the following idea:
a=3
. <(echo a=4; echo b=5)
echo $a
echo $b
It allows returning different variables at once (which could be an issue in the subshell variant of the accepted answer).
No iteration is needed,
No temporary file to take care of.
Close to the syntax proposed by the OP.
Result:
4
5
With xtrace enabled is visible that we are sourcing from the file descriptor created for the output of the subshell:
+ a=3
+ . /dev/fd/63 # <-- the file descriptor ;)
++ echo a=4
++ echo b=5
++ a=4
++ b=5
+ echo 4
4
+ echo 5
5
You can output the value in the subshell and assign the subshell output to a variable in the caller script:
# subshell.sh
echo Value
# caller
myvar=$(subshell.sh)
If the subshell has more to output you can separate the variable value and other messages by redirecting them into different output streams:
# subshell.sh
echo "Writing value" 1>&2
echo Value
# caller
myvar=$(subshell.sh 2>/dev/null) # or to somewhere else
echo $myvar
Alternatively, you can output variable assignments in the subshell, evaluate them in the caller script and avoid using files to exchange information:
# subshell.sh
echo "a=4"
# caller
# export $(subshell.sh) would be more secure, since export accepts name=value only.
eval $(subshell.sh)
echo $a
The last way I can think of is to use exit codes but this covers the integer values exchange only (and in a limited range) and breaks the convention for interpreting exit codes (0 for success non-0 for everything else).
Instead of accessing the variable from the parent shell, change the order of the commands and use the process substitution:
a=3
echo 5 | (read a)
echo $a
prints 3
a=3
read a < <(echo 5)
echo $a
prints 5
Another example:
let i=0
seq $RANDOM | while read r
do
let i=r
done
echo $i
vs
let i=0
while read r
do
let i=r
done < <(seq $RANDOM)
echo $i
Alternatively, when job control is inactive (e.g. in scripts) you can use the lastpipe shell option to achieve the same result without changing the order of the commands:
#!/bin/bash
shopt -s lastpipe
let i=0
seq $RANDOM | while read r
do
let i=r
done
echo $i
Unless you can apply all io to pipes and use file handles, basic variable updating is impossible within $(command) and any other sub-process.
Regular files, however, are bash's global variables for normal sequential processing. Note: Due to race conditions, this simple approach is not good for parallel processing.
Create an set/get/default function like this:
globalVariable() { # NEW-VALUE
# set/get/default globalVariable
if [ 0 = "$#" ]; then
# new value not given -- echo the value
[ -e "$aRam/globalVariable" ] \
&& cat "$aRam/globalVariable" \
|| printf "default-value-here"
else
# new value given -- set the value
printf "%s" "$1" > "$aRam/globalVariable"
fi
}
"$aRam" is the directory where values are stored. I like it to be a ram disk for speed and volatility:
aRam="$(mktemp -td $(basename "$0").XXX)" # temporary directory
mount -t tmpfs ramdisk "$aRam" # mount the ram disk there
trap "umount "$aRam" && rm -rf "$aRam"" EXIT # auto-eject
To read the value:
v="$(globalVariable)" # or part of any command
To set the value:
globalVariable newValue # newValue will be written to file
To unset the value:
rm -f "$aRam/globalVariable"
The only real reason for the access function is to apply a default value because cat will error given a non-existent file. It is also useful to apply other get/set logic. Otherwise, it would not be needed at all.
An ugly read method avoiding cat's non-existent file error:
v="$(cat "$aRam/globalVariable 2>/dev/null")"
A cool feature of this mess is that you can open another terminal and examine the contents of the files while the program is running.
While it's harder to get multiple variables out of a subshell, you can set multiple variables inside a function without using globals.
You can pass the name of a variable into a function that uses local -n to turn it into a special variable called a nameref:
myfunc() {
local -n OUT=$1
local -n SIDEEFFECT=$2
OUT='foo'
SIDEEFFECT='bar'
}
myfunc A B
echo $A
> foo
echo $B
> bar
This is the technique I ended up using instead of getting subshell FOO=$(myfunc) working setting multiple variables.
A very simple and practical method that allows multiple variables is as follows, eventually may add parameters to the call:
function ComplexReturn(){
# do your processing...
a=123
b=456
echo -n "AAA=${a}; BBB=${b};"
}
# ... this can be internal function or any subshell command
eval $(ComplexReturn)
echo $AAA $BBB