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Setting a variable within if block

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# 1  
Old 08-18-2009
Setting a variable within if block
Hi,

i have a variable which i would like to set inside an if block

for example
Code:
IS_VAR=0
if [ "true" = "true" ]
then
 IS_VAR=1
fi
echo IS_VAR

the last echo statement gives 0.So setting variables in the if block doesnt have effect outside the block?Is there any workaround for this?

Thanks ,
Padmini
Last edited by Franklin52; 08-18-2009 at 10:01 AM.. Reason: Please use code tags!
# 2  
Old 08-18-2009
nothing like that.. $ is missing from your last echo statement
Code:
home/vidya_perl> cat vv
IS_VAR=0
if [ "true" = "true" ]
then
IS_VAR=1
fi
echo $IS_VAR
home/vidya_perl> vv
1

# 3  
Old 08-18-2009
Hi,

i use echo $IS_VAR only.

and the example is slightly modified here

its like,
I
Code:
S_VAR=0
ls | while read input
do
if [ $input = "hi" ]
then
  IS_VAR=1
  echo $IS_VAR
  break
fi
done
echo $IS_VAR

for the first echo , i get 1 and for the second i get 0.
Last edited by Franklin52; 08-18-2009 at 10:02 AM.. Reason: Please use code tags!
# 4  
Old 08-18-2009
Quote:
Originally Posted by padmisri
Hi,

i use echo $IS_VAR only.

and the example is slightly modified here

its like,
IS_VAR=0
ls | while read input
do
if [ $input = "hi" ]
then
IS_VAR=1
echo $IS_VAR
break
fi
done
echo $IS_VAR

for the first echo , i get 1 and for the second i get 0.
Yes,looks the initialization inside the while block has no effect outside the while block.
But if you remove while then output 1 and 1.
Code:
#!/bin/sh

IS_VAR=5
input="hi"

if [ $input = "hi" ];then    
    IS_VAR=1
    echo $IS_VAR
fi

echo $IS_VAR

# 5  
Old 08-18-2009
The pipe starts a new shell (child) so the assigned value of the variable is lost after the loop.

Try something like this:

Code:
IS_VAR=0

list=$(ls)

for input in $list
do
if [ $input = "hi" ]
then
  IS_VAR=1
  echo $IS_VAR
  break
fi
done
echo $IS_VAR

Regards
# 6  
Old 08-18-2009
Quote:
Originally Posted by Franklin52
The pipe starts a new shell (child)
The script does work for me...

Code:
> cat Test
#!/bin/sh

touch hi
echo Outside $PPID $$
ls | while read input
do
if [ $input = "hi" ]
then
  echo Inside $PPID $$
  IS_VAR=1
  echo Inside $IS_VAR
  break
fi
done
  echo Output $PPID $$
echo Outside $IS_VAR
> ./Test
Outside 13426702 2252968
Inside 13426702 2252968
Inside 1
Output 13426702 2252968
Outside 1

Which OS are you using, Padmini?

Edit: Wow... It works on AIX but not on Linux or Solaris. I do apologise.
Last edited by Scott; 08-18-2009 at 10:37 AM..
# 7  
Old 08-18-2009
It doesn't work in ubuntu.
This is the output:
Code:
Outside 9559 9803
Inside 9559 9803
Inside 1
Output 9559 9803
Outside

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