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Re: I have a folder which is having 1200 files,all the file names are have maximum of 8 character length.I need to remove last character from each of the filnames and prefix it with character "A"

by MidLifeXis (Prior)
on Feb 20, 2013 at 18:38 UTC ( #1019822=note: print w/ replies, xml ) Need Help??


in reply to I have a folder which is having 1200 files,all the file names are have maximum of 8 character length.I need to remove last character from each of the filnames and prefix it with character "A"

See rename, split, join, length, and substr for some ideas. You may also want to see -e to check if a file or directory exists. Before using rename, you will likely want to do a test run with just printing.

--MidLifeXis


Comment on Re: I have a folder which is having 1200 files,all the file names are have maximum of 8 character length.I need to remove last character from each of the filnames and prefix it with character "A"
Re^2: I have a folder which is having 1200 files,all the file names are have maximum of 8 character length.I need to remove last character from each of the filnames and prefix it with character "A"
by perladdict (Chaplain) on Feb 20, 2013 at 20:38 UTC
    Thanks MidLifeXis, As per your inputs i have tried to implement this,Below is the code i tried it,i am able to succeed to rename all tons of files in my directory.
    #!/usr/bin/perl $dir = "C:\\Users\\user\\Desktop\\test"; opendir DIR, $dir or die "error: cannot open directory \"$dir\" $!"; @files = sort grep (!/^\.$|^\.\.$/, readdir(DIR)); closedir (DIR); foreach $fl(@files) { if($fl =~ m/\.doc/) { @a = split(/\.doc/,$fl); foreach $len(@a) { $z = length $len; if($len <= $z) { @x = substr($len,0,7); foreach $mod(@x) { $mod = "A".$mod.".doc"; rename $fl,$mod; } } } } }

      Grep pattern used for getting files can be modified to reduce the for loop iteration as you're going to rename only file names which are less than or equal to 8 characters with .doc extension.

      @files = sort grep /^.{1,8}\.doc$/, readdir(DIR);

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