UNIX for Dummies Questions & Answers

I am new to shell scripting
can u guys please provide a small script for the following senario

step1:need to find some files in a directory for ex having 020908
step2:sort them and redirecting to new file
(ex:sort abc > abc.sort)

i am trying this but giveing flag error
ls -l *.020908* |grep -v .sort| while read file
do
sort "$file" > "$file".sort
done

Cheers
Reddy

The ls -l is Useless and parsing the long-format output is troublesome.

Code:

for f in *.020908*; do
  case $f in *.sort) continue ;; esac
  sort "$file" >"$file".sort
done

ths era

while executing this script i got this error

for f in *.020908*; do
case $f in *.sort) continue ;; esac
sort "$file" >"$file".sort
done

Cannot open %s

In your original script "ls -l" (ell) should have been "ls -1" (one) to just get the filename.

Here's a version with some minor tweaks. It starts with ls hyphen one.

ls -1 *\.020908* 2>/dev/null | grep -v "\.sort$" | while read file1
do
file2="${file1}.sort"
echo "Sorting ${file1} into ${file2}"
sort < "${file1}" > "${file2}"
done

Thx ! its working
could you please expalin why do we need


2>/dev/null


and this logic
file2="${file1}.sort"
echo "Sorting ${file1} into ${file2}"
sort < "${file1}" > "${file2}"