shell programming

sajith · 12-05-2005, 09:21 AM

hai sir

i have a doubt in shell programming

when i do shell program

in that declare a variable count=0

and use this variable inside a loop and incremented each when specified condition of the loop is satisfied

and last print the value of variable oustide the loop

at this it display the value of variable as 0(zero,initialization value)

but inside the loop the incremented each time
but when print the final value it will not display it correctly

sir please give me solution for this problem

crabboy · 12-05-2005, 09:26 AM

I'd have to see a script to comment on your problems, but the script below works as you describe.

Code:

#!/bin/sh

COUNT=0

while [ $COUNT -lt 5 ]; do
   echo $COUNT
   COUNT=`expr $COUNT + 1`

done

echo "Fishished: $COUNT"

Code:

# ./loop.sh
0
1
2
3
4
Fishished: 5

sajith · 12-05-2005, 09:40 AM

hai sir

here i am give the code

but it will not work properly

sir's code is working
but i implement the same thing inside a while loop and if statement

but it will not give the correct final value of the variable

code is

#!/bin/bash
count=0
ls -lS $1>tmp
echo -e filename \\t filesize

tail +2 tmp|while read line
do
f_size=`echo $line|cut -f5 -d" " `
if [ $f_size -gt 1000 ]
then

echo -e ` echo $line|cut -f9 -d" " ` \\t ` echo $line|cut -f5 -d" " `
count=` expr $count + 1 `
echo $count

fi

done

crabboy · 12-05-2005, 09:45 AM

You are not looping based on the condition of 'count'. You are looping based on what is read from a file. So when the count condition is met the loop may/may not continue.

I'd suggest not looping on the read and do something similar to what I've previously posted.

LasseW · 12-05-2005, 10:53 AM

Hi Sajith,

The problem is that the pipe will spawn a new shell and the count increment is local to that shell. Write the while loop like this instead:

while
...
done <tmp

To get rid of the total line before the loop:

ls -lS $1 | tail +2 >tmp