LinuxQuestions.org - How to define a variable In a while loop?

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How to define a variable In a while loop?

I'm writing a bash package management script and I have a variable, $conflict, that is set to "1" if a system-package conflict exists (the system contains the same files that the package does). Here's the loop where the script checks for conflicts:

Code:

#### CHECK FOR PACKAGE - SYSTEM CONFLICTS

#### Set $conflict to "1" if a conflict exists.



find | \

# Outputs all files and directories under the PWD.

sed "/\.$/d" | \

# Deletes the current directory (.) entry.

sed "s@^.@@" | \

# Removes the "." char from the beginning of all the file paths



while read file ; do

echo 'Entering Loop #1'



#echo "\$file = $file"



sysFile=`echo $file | sed "s@^@$sysRoot@"`

pkgFile=`echo $file | sed "s@^@$pkgPath@"`



#echo '$sysFile =' $sysFile

#echo '$pkgFile =' $pkgFile



if [ -d $pkgFile ] && [ -e $sysFile ] && ! [ -d $sysFile ] ; then



        echo "ERROR: $sysFile exists!"

        conflict="1"

        break



fi





if ! [ -d $pkgFile ] && [ -e $sysFile ] ; then

                

        echo "ERROR: $sysFile exists!"

        conflict="1"

        break

        

fi



echo 'Exiting Loop #1'



echo $conflict

done

My problem is that when I set $conflict to "1", it only remains "1" within the loop. Right after the break command, it returns to the default value which I defined at the beginning of the script: "0". How would I make a global change to the variable (meaning that it will = "1" throughout the script)?

Shouldn't the
echo $conflict
be outside of the while loop ?

Quote:

Originally Posted by dxqcanada

Shouldn't the
echo $conflict
be outside of the while loop ?

There's one outside of the loop as well, I just didn't include that segment.

Since you are breaking out of the loop the echo $conflict that is in the loop will show "0" (since conflict only get set to 1 before breaking out of the loop.

The echo $conflict that is outside of the loop should show "1" if the condition is met.

This simple script will show you"

Code:

#!/bin/bash

conflict=0

while read input

do

if [ $input = 0 ]

then

conflict=1

break

fi

echo conflict inside is $conflict

done

echo conflict outside is $confict

Perhaps try to "define" the variable before the loop?

ya I guess, just do conflict=0 before beginning of while loop.

Quote:

Originally Posted by dxqcanada

Code:

#!/bin/bash

conflict=0

while read input

do

if [ $input = 0 ]

then

conflict=1

break

fi

echo conflict inside is $conflict

done

echo conflict outside is $confict

That example returned errors... Not sure why but here's my edited code:

Code:

#!/bin/bash



# $conflict will be set to "1" if conflict exists

conflict="0"

echo $conflict



sysRoot="/home/ubuntu/Desktop/sources/sysRoot"

rootPkgDir="/home/ubuntu/Desktop/sources/pkg"

pkg="lynx-2.8.6"

pkgPath="$rootPkgDir/$pkg"

#pkgPath="/home/ubuntu/Desktop/sources/pkg/lynx-2.8.6"



echo '$rootPkgDir = ' $rootPkgDir

echo '$pkg = ' $pkg

echo '$pkgPath = ' $pkgPath



cd $pkgPath









#### CHECK FOR PACKAGE - SYSTEM CONFLICTS

#### Set $conflict to "1" if a conflict exists.



find | \

# Outputs all files and directories under the PWD.

sed "/\.$/d" | \

# Deletes the current directory (.) entry.

sed "s@^.@@" | \

# Removes the "." char from the beginning of all the file paths



while read file ; do

#echo 'Entering Loop #1'



#echo "\$file = $file"



sysFile=`echo $file | sed "s@^@$sysRoot@"`

pkgFile=`echo $file | sed "s@^@$pkgPath@"`



#echo '$sysFile =' $sysFile

#echo '$pkgFile =' $pkgFile



if [ -d $pkgFile ] && [ -e $sysFile ] && [ ! -d $sysFile ] ; then



        echo "ERROR#1: $sysFile exists."

        conflict="1"

        echo "Setting \$conflict to \"1\""

        echo $conflict

        break



fi





if [ ! -d $pkgFile ] && [ -e $sysFile ] ; then

                

        echo "ERROR#2: $sysFile exists."

        conflict="1"

        echo "Setting \$conflict to \"1\""

        echo $conflict

        break

        

fi



#echo 'Exiting Loop #1'



done



echo '$conflict =' $conflict "outside the loop."

Here's the output when I run the script:

Quote:

0
$rootPkgDir = /home/ubuntu/Desktop/sources/pkg
$pkg = lynx-2.8.6
$pkgPath = /home/ubuntu/Desktop/sources/pkg/lynx-2.8.6
ERROR#2: /home/ubuntu/Desktop/sources/sysRoot/usr/bin/lynx exists.
Setting $conflict to "1"
1
$conflict = 0 outside the loop.

as you can see, the second if statement evaluates as true ( I have set up a conflict on purpose to test it). Then $conflict is set to "1". I echo conflict inside the if statement, it returns the value "1". Then I issue the break command to exit the loop. Outside the loop, I echo $conflict, this time it evaluates as "0". Why?

it seems to me that variable keeps its value within the loop...
try "export $conflict " and echo $conflict ( to see the value)... inside the loop

The problem is the pipe. When you pipe a new instance of the shell is executed as a separate process to receive data on stdin. When you make a variable assignment inside this new pipe process the variable is lost when the process exits at the end of the pipe.

Quote:

My problem is that when I set $conflict to "1", it only remains "1" within the loop. Right after the break command, it returns to the default value which I defined at the beginning of the script: "0". How would I make a global change to the variable (meaning that it will = "1" throughout the script)?

try declare -r conflict=1

"
-r Make variables read-only. These variables cannot then be assigned values by subsequent assignment statements, nor can they be unset.
"

Here's a short example that demonstrates your situation.

Code:

#!/bin/bash



linenum=0

echo -en "hello\nworld\n" | while read x; do

    linenum=$((linenum+1))

    echo "$linenum: $x"

done

echo "total number of lines = $linenum"



It then prints out...

1: hello

2: world

total number of lines = 0

The problem is you want the total number of lines to be printed as 2. This can not be done with a variable assignment in the loop because the pipe ( | while read ...) runs in a separate process.

A workaround is to put the entire "stuff | while blah; do halb; done" construct inside backquotes and echo $conflict at the end. i.e.

Code:

conflict=`stuff | ( while blah; do whatever; done; echo $conflict )`

Be sure to echo any errors found inside the subshell to stderr like

Code:

echo >&2 "Oh no Mr. Bill!"

so that they aren't captured by the backquotes

We have the same problem about variable scoping, since I also need to get the presently login user and then, check if corresponding user belongs to a group.

I have wrote several test code with regards to variable scoping outside while loop. But, I could not find any related variable declaration to make the variable value within WHILE loop, I also use export and using extra conditions. The result leads me to conclude that variable within While Loop is only accessible from within the loop itself.

So, I decided to save the value to a file then read it back outside the while loop -- it resolves my problem, please see below code if you could find it useful:

whoami | while read line; do
#if [[ "$userName" = "" ]] ; then
userName=$line
#fi
echo $userName
if [ ! $userName == "" ]; then
echo "userName=$userName" > ~/getuser.bash
break
fi
done
chmod 755 ~/getuser.bash
echo "username value outside while loop=$userName"

if [ -f getuser.bash ] ; then
source getuser.bash
fi
echo "username value outside while loop in getuser.bash file=$userName"

Output:
username value outside while loop=
username value outside while loop in getuser.bash file=hpal

@hpal -- You are responding to a thread that is over 5 years old. The OP either solved his problem or gave up long ago.

or dead.
but FYI use export:

Code:

$ export x=fish

$ echo $x

fish

$ while :;do x=chips;break;done

$ echo $x

chips

export passes variables to subprocesses, but that doesn't help getting the data back to the parent:

Code:

$ export x=fish

$ echo $x

fish

$ whoami | while :;do x=chips;break;done

$ echo $x

fish

The problem is explained in detail here: 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?

Another way to fix the problem is to place the input into a file,
then redirection to have the while loop process the file

i.e.

-----------------------------------------------------------
Doesn't work due to pipe causing subprocess being created:
-----------------------------------------------------------
export conflict=0
cat $filename | while read inline
do
conflict=1
done

echo $conflict # get 0

--------------------
WORKS!
--------------------
export conflict=0
while read inline
do
conflict=1
done < $filename

echo $conflict # get 1 !

Since no pipe, no subprocess created, values in while loop retained