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 "$var" #second echo
The reason for this behaviour is that a while loop runs in a subshell when it's part of a pipeline. For the while loop above, a new subshell with its own copy of the variable var is created.
See this article for possible workarounds: I set variables in a loop that's in a pipeline. Why do they disappear after the loop terminates? Or, why can't I pipe data to read?.
Related
Bash allows to use: cat <(echo "$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
}
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 was trying to understand how "while read" works in bash, and i came accross a behaviour that i can't explain:
root#antec:/# var=old_value
root#antec:/# echo "new_value" | read var && echo $var
old_value
root#antec:/#
It works fine with "while read":
root#antec:/# var=old_value
root#antec:/# echo "new_value" | while read var; do echo $var; done
new_value
root#antec:/#
Can someone explain why it does not work when read is used without while ? Moreover, i don't understand how the value of "var" in the main shell can be seen from the allegedly subshell after the pipe ..
Thank you
I believe this is precedence problem, with | having higher precedence than &&. First is grouped as:
(echo "new_value" | read var) && echo $var
Second is grouped as:
echo "new_value" | (while read var; do echo $var; done)
Why do you think it works in the second case? It doesn't really update var. do echo $var after the line with while and see.
Explanation: whatever comes after | is executed in a subshell that has its own copy of var. The original var is not affected in either case. What is echoed depends on whether echo is called in the same subshell that does read or not.
I didn't want to print the variable immediately but use it later on and this works:
var=old_value
read var < <(echo "new_value")
echo $var
> new_value
alternatively:
var=old_value
tmp=$(echo "new_value")
read var <<<"$tmp"
echo $var
> new_value
You don't need a while loop:
echo "new_value" | (read var; echo $var)
i have bash script such as
for i in `echo a [ matched.lines`
do
echo $i
done
why output of this script below
a
[
matched.lines
i want output text as it is
a [ matched.lines
how can i do that
thanks for help
The script echos each token seperately. Use echo -n $i inside do ... done.
i have replaced ' ' characters to '' with sed global replace and there is no new line on output of script thanks for help
to read a file using shell, use while read loop
while read -r loop
do
case "$line" in
*]* ) echo $line;;
esac
done <"file"
Sorry but that's not going to do what you want it to. Your for loop is using a space as a delimiter to it's arguments. This will cause each one of the item's to echo on a seperate line. Adding a -n doesn't work here because you will not get the \n at the end of the line. Honestly I don't see what it is your trying to do here. If you just want to echo this "a [ matched.lines" as a string you can do this:
for i in "a [ matched.lines";do
echo $i
done
But I feel you're going to misunderstand how to use a for loop....