Shell Programming and Scripting

hi ,

could any one tell me why i m not getting the value of variable i outside the
while loop in file reading

cat filename | while read line
do
i=2
done

echo $i

here it is not showing the value of i . what i have to do if i want to persist the value of
i outside the while loop



thanks

It currently doesn't show the value of `i` because it never reaches it.
filename has to contain something for the loop to be entered at least once.

Otherwise it works fine.

Cheers,
Cameron

no filename contains some names like

ram
ravi



so it will go 2 times in the loop . even though it is not working

In this case, the pipe spawns a sub-shell and assigns i=2. When it comes out of the subshell, it does retain the value of i. Think scoping and you will understand.

Do this and you will get what you are looking for

Code:

while read line
do
  i=2
done < filename
echo $i

My reasoning:

Code:

[mddev:/home/cameron]
$ cat detail2.sh
#!/bin/ksh

cat /home/cameron/dummyfile.txt | while read inline
do
  i=2
done

echo $i


[mddev:/home/cameron]
$ detail2.sh
2

[mddev:/home/cameron]
$

Nice pickup Vino - not seen that before (not that I can recall).
What system/application settings cause this type of result?

Cheers,
Cameron

It's the pipe that puts the "while/read/do/done" in a child process.

If you did

while read N; do xxxx ; done <FILE

it would not go in a child process.

Hi porter,
I understand that the child process being spawned.
But why is it that when I run it, `i` is resolved; and when useless79 ran the same `i` didn't resolve ??

Is it a shell attribute ??

Sorry for the continued questions - just very curious to understand.

Cheers,
Cameron