condition inside a for loop

Shell Programming and Scripting
1

I have a for loop in my script as shown below.

for file_path in $file_list ; do

........my code
..........
......

done

Can i restrict the number of files parsing to the variable file_path as 50?
That is, even if I have some 100 files in file_list, I need to take only 50 files for the loop every time.

Is that possible? Can someone help me with this?

Thanks,
Vijay

2 
for file_path in `cat filelist.txt|head -50`
do
        echo $file_path
--
---
YOUR_CODE_HERE
----
done
3 
head -50 file | while read file_path
do
 .......
done
4

But.. this doesn't solve my request.
Because.. i read it from a variable $file_list.
This contains the input as following:

echo $file_list
aaa.txt bbb.txt ccc.txt ddd.txt .......and so on.

So all file names will be in a single line. so head -50 will return all the files.. right?

any other way to get the first 50 names in the variable?

5

Just maintain a counter in the loop, and exit when it reaches 50.

n=0
for file_path in $file_list; do
  : : : stuff
  n=`expr $n + 1`
  case $n in 50) break;; esac
done
6 
filelist="1 2 3 4 5 6 7 8 9 10"
for filepath in `echo $filelist|cut -d" " -f1-5`
do
        echo $filepath
done
 
OUTPUT
1
2
3
4
5

Instead of 5 use 50.

7

Assuming that spacing in the variable is regular. If it might contain tabs or newlines, awk might be a better solution. Or maybe

perl -le '$,=" "; print @ARGV[0..49]' "$@"
8

Maybe this will help,

echo "$file_list" | awk '{ for(i=1;i<51;i++) print $i }' | while read file
   
           do
                   # actions with "$file"
           done