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 variable:
$var="-- ---comment- --
abcd;"
I want to remove every character that is between "--" comment "--"
Expected output:
$var="abcd;"
What I am trying in test.sh
#!/bin/bash
var="----comment---
abcd;"
test=$(sed 's/^--.*--/' $var)
echo $test
It is not working. I get the following error:
command not found
Trying to replace the pattern that you don't expect in output with nothing, something like this
test=$(echo $var | sed 's/^--.*--//g')
Should work.
In bash:
$ cat foo.sh
var="----comment---
abcd;"
echo "$var" # this outputs the original var
echo "${var/--* /}" # this output replacing everything between -- and " "
$ bash foo.sh
----comment---
abcd;
abcd;
You can set the output to a variable by
$ var="${var/--* /}"
I have two bash string built in commands that work fine independently but when nested generate an error message no matter what I try. Here's the two individual commands that work:
$ A="etc/.java"
$ echo $A
/etc/.java
$ B="${A//$'\057\056'/$'\057'}"
$ echo $B
/etc/java
$ B="${A^^}"
$ echo $B
/ETC/.JAVA
Now trying to combine the two commands together I get errors:
$ B="${${A^^}//$'\057\056'/$'\057'}"
bash: ${${A^^}///.//}: bad substitution
$ B="${ ${A^^}//$'\057\056'/$'\057'}"
bash: ${ ${A^^}///.//}: bad substitution
$ B="${ ${A^^} //$'\057\056'/$'\057'}"
bash: ${ ${A^^} ///.//}: bad substitution
$ B="${"${A^^}"//$'\057\056'/$'\057'}"
bash: ${"${A^^}"//'/.'/'/'}: bad substitution
$ B="${ "${A^^}" //$'\057\056'/$'\057'}"
bash: ${ "${A^^}" //'/.'/'/'}: bad substitution
$ B="${${A^^} //$'\057\056'/$'\057'}"
bash: ${${A^^} ///.//}: bad substitution
Simplified examples are presented above so one can copy and paste to their own terminal. Piping or redirection would be complicated because my real world code is this:
while [[ $i -lt $DirsArrCnt ]] ; do
DirsArr[$i]=false
CurrNdx=$i
CurrKey="${DirsArr[$(( $i + 1 ))]}"
# ^^ = convert to upper-case
# ${Variable//$'\041\056'/$'\041'} = Change /. to / for hidden directory sorting
if [[ "${"${CurrKey^^}"//$'\041\056'/$'\041'}" > \
"${"${LastKey^^}"//$'\041\056'/$'\041'}" ]] || \
[[ "${"${CurrKey^^}"//$'\041\056'/$'\041'}" = \
"${"${LastKey^^}"//$'\041\056'/$'\041'}" ]] ; then
LastNdx=$CurrNdx
LastKey="$CurrKey"
i=$(( $i + $OneDirArrCnt))
continue
fi
In the special case of one of the expansions being upper casing, it can be done in a single expansion, using declare -u (introduced in Bash 4.0). declare -u converts to uppercase on assignment.
Combining upper casing and substitution then becomes this:
$ declare -u A='/etc/.java'
$ echo "${A//\/./\/}"
/ETC/JAVA
There is the analogous -l for lower casing and the (undocumented) -c for title casing, but these are the only cases where you can do "nested" parameter expansion.
Let's say I have two variables:
a="AAA"
b="BBB"
I read a string from a file. This string is the following:
str='$a $b'
How to create a new string from the first one that substitutes the variables?
newstr="AAA BBB"
bash variable indirection whithout eval:
Well, as eval is evil, we may try to make this whithout them, by using indirection in variable names.
a="AAA"
b="BBB"
str='$a $b'
newstr=()
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=($cnt)
done
newstr="${newstr[*]}"
echo $newstr
AAA BBB
Another try:
var1="Hello"
var2="2015"
str='$var1 world! Happy new year $var2'
newstr=()
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=($cnt)
done
newstr="${newstr[*]}"
echo $newstr
Hello world! Happy new year 2015
Addendum As correctly pointed by #EtanReisner's comment, if your string do contain some * or other glob expendable stings, you may have to use set -f to prevent bad things:
cd /bin
var1="Hello"
var2="star"
var3="*"
str='$var1 this string contain a $var2 as $var3 *'
newstr=()
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt};
newstr+=("$cnt");
done;
newstr="${newstr[*]}"
echo "$newstr"
Hello this string contain a star as * bash bunzip2 busybox....zmore znew
echo ${#newstr}
1239
Note: I've added " at newstr+=("$cnt"); to prevent glob expansion, but set -f seem required...
newstr=()
set -f
for cnt in $str ;do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=("$cnt")
done
set +f
newstr="${newstr[*]}"
echo "$newstr"
Hello this string contain a star as * *
Nota 2: This is far away from a perfect solution. For sample if string do contain ponctuation, this won't work again... Example:
str='$var1, this string contain a $var2 as $var3: *'
with same variables as previous run will render:
' this string contain a star as *' because ${!var1,} and ${!var3:} don't exist.
... and if $str do contain special chars:
As #godblessfq asked:
If str contains a line break, how do I do the substitution and preserve the newline in the output?
So this is not robust as every indirected variable must be first, last or space separated from all special chars!
str=$'$var1 world!\n... 2nd line...'
var1=Hello
newstr=()
set -f
IFS=' ' read -d$'\377' -ra array <<<"$str"
for cnt in "${array[#]}";do
[ "${cnt:0:1}" == '$' ] && cnt=${cnt:1} && cnt=${!cnt}
newstr+=("$cnt")
done
set +f
newstr="${newstr[*]}"
echo "$newstr"
Hello world!
... 2nd line...
As <<< inline string add a trailing newline, last echo command could be written:
echo "${newstr%$'\n'}"
The easiest solution is to use eval:
eval echo "$str"
To assign it to a variable, use command substitution:
replaced=$(eval echo "$str")
Disclaimer: I only discovered perl an hour ago. But this seems to work robustly, whatever special characters you throw at it:
newstr=$(a2="$a" b2="$b" perl -pe 's/\$a\b/$ENV{a2}/g; s/\$b\b/$ENV{b2}/g' <(echo -e "$str"))
Test:
a='A*A\nA'
b='B*B\nB'
str='$a $aa * \n $b $bb'
newstr=$(a2="$a" b2="$b" perl -pe 's/\$a\b/$ENV{a2}/g; s/\$b\b/$ENV{b2}/g' <(echo -e "$str"))
echo -e "$newstr"
Output:
A*A
A $aa *
B*B
B $bb
I'd use awk solution with awk-variables. This will allow passing a text containing special chars and subsitute any placeholder with it.
a workaround to recognize $ would be using [\x24]:
awk -v a="$a" -v b="$b" '{gsub("[\x24]a",a);gsub("[\x24]b",b); print}' <<< $str
here
-v defines variable a="$a"
[x24] is ASCII for $, so [x24]a equal to $a
gsub(x,y) - replaces x with y
I want to print string called "$1". But when I do this with echo it prints string which equals to "$1" variable. How can I print "$1" just like string?
for example:
set -- "output" # this sets $1 to be "output"
echo $1 # ==> output
But I want this:
echo $1 # ==> $1
You have to escape the $ to have it shown:
$ echo "\$1"
or, as noted by JeremyP in comments, just use single quotes so that the value of $1 does not get expanded:
$ echo '$1'
You need to either:
Enclose the variable in SINGLE quotes: echo '$1'
Escape the $ sign: echo "\$1"