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Full Discussion: Renaming file in batch
Top Forums Shell Programming and Scripting Renaming file in batch Post 302929659 by Don Cragun on Wednesday 24th of December 2014 03:53:56 AM
Old 12-24-2014
Quote:
Originally Posted by RavinderSingh13
Hello mfaizan40,

... ... ...

Code:
cat move_files.ksh
default=92311
for j in "bep01.smsc.*"
do
        A=$j
done
for i in $A
do
        value=`echo $i | awk '{gsub(/\-.*/,X,$0);print}'`
        new_value=$value"-"$default"".cdr"
        mv $i $new_value
        echo file $i " has been renamed to " $value"-"$default".cdr"
        let "default = default + 1"
done

... ... ...
Ravinder correctly notes that expanding the parameter list in a for statement avoids the possible argument list too long problem.

But:
  1. I don't understand the need for the 1st for loop.
  2. Calling awk 500 times seems like a lot of overhead when parameter substitution can be done in the shell without the overhead of invoking a separate process.
  3. There is a mismatched double quote in $value"-"$default"".cdr". Perhaps "$value-$default.cdr" is what was intended.
  4. While just expanding $default works fine as long as it starts out as a five digit number, it might be safer to use printf with a five digit leading zero filled format to cover the general case.
  5. It seems to be important to match the terminating -00000.cdr at the ends of the filenames in case other files starting with bep01.smsc. have already been renumbered.
  6. The let command is less portable than arithmetic substitution.

Taking the above considerations into account, you might want to also try something like:
Code:
#!/bin/ksh
sequence=92311
for i in bep01.smsc.*-00000.cdr
do	new="${i%00000.cdr}$(printf '%05d.cdr' "$sequence")"
	# mv $i $new
	echo "file $i has been renamed to $new"
	sequence=$((sequence + 1))
done

which will work with ksh, bash, or any other shell that performs POSIX standard parameter expansions and arithmetic substitutions.

If the echo statements show that the renaming calculations are correct, remove the # from the start of the mv command line so the script will actually move the files.
This User Gave Thanks to Don Cragun For This Post:
 

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