Global Variable in shell script

Question

I would like to make z a global variable in the following code:

#!/bin/bash                                                                                                                          
z=0;
find $1 -name "*.txt" | \
while read file
do
  i=1;
  z=`expr $i + $z`;
  echo "$z";
done
echo "$z";

The last statement always outputs "0". Why?

You could use z=$(( z + i )) or (( z += i )) to about calling out to expr — glenn jackman
– glenn jackman, Commented Oct 25, 2011 at 13:25

Ignacio Vazquez-Abrams · Accepted Answer · 2011-10-25 07:07:58Z

5

Pipes start a new subshell.

answered Oct 25, 2011 at 7:07

Ignacio Vazquez-Abrams

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1 Comment

Vivek Goel Over a year ago

so how can I get last value inside z ?

glenn jackman · Accepted Answer · 2011-10-25 13:27:39Z

1

The simple way to translate

find ...  | while read ...; done

to a form without pipes is using process substitution:

while read ...; done < <(find ...)

Readability suffers somewhat.

answered Oct 25, 2011 at 13:27

glenn jackman

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Comments

Luo · Accepted Answer · 2011-10-25 07:12:33Z

0

I don't know why this happened, but the problem is caused by the pipe.

If you do it like this

#!/bin/bash                                                                                                                          
    z=0;
    for f in `find $1 -name "*.txt"`
    do
    i=1;
    z=`expr $i + $z`;
    echo "$z";
    done
    echo "$z";

then $z will not be zero.

answered Oct 25, 2011 at 7:12

Luo

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