I am new to scripting, and I have a directory with all files named num.pdb.ostat. I would like to rename all num.ostat (that is deleting the .pdb in all). For a single file this works:

mv 2.pdb.ostat 2.ostat

but when I try to do it for all files in folder with this script

for num in ./*; do mv ${num}.pdb.ostat ${num}.ostat; done

nothing happens

Can anyone tell me, where I went wrong?

  • 1
    No. But ever seen the rename.ul utility? It will be something like this: rename.ul pdb. . *ostat and it should rename the files.
    – Rinzwind
    Mar 9, 2015 at 9:17
  • Or (assuming you are using perl rename) rename 's/pdb.//' *.pdb.ostat
    – Sparhawk
    Mar 9, 2015 at 9:19
  • Does that mean, I can bypass the loop and just use the rename.ul when I am in the directory? It gives me this response : -bash: syntax error near unexpected token `pdb.'
    – Ditte
    Mar 9, 2015 at 9:20
  • @akabhirav The question is Can anyone tell me, where I went wrong?, so it's not a dupe exactly, although the link is certainly helpful.
    – Sparhawk
    Mar 9, 2015 at 9:29

4 Answers 4


${num} takes the whole file name. You need to get filename without extension and add your new extension. You can make a string formatting. Use the following command:

for num in ./*; do mv ${num} ${num%.*.*}.ostat ; done

% deletes shortest match of $substring from back of $string.

  • 1
    Thank you. This was so simple that I could understand it, and it worked nicely :)
    – Ditte
    Mar 9, 2015 at 10:54
  • 2
    +1 I always forget about parameter substitution. This won't work for files with spaces in them, but is fine for this question. Also, you could use the greedy expression ${num%%.*}.ostat instead, and save one character! :)
    – Sparhawk
    Mar 11, 2015 at 1:42

For troubleshooting your script, try replacing mv with echo. You'll see that your variable ${num} contains the full file name, e.g. 2.pdb.ostat. Hence your script essentially tries to run

mv ./2.pdb.ostat.pdb.ostat ./2.pdb.ostat.ostat

Instead, you have to truncate the filename back to just the number first. e.g.

for filename in ./*; do num="$(echo "${filename}" | grep -o '^./[0-9]*')"; echo "${num}.pdb.ostat" "${num}.ostat"; done

Once you have confirmed the syntax is okay, you can change the echo to a mv, and actually move the files. i.e.

for filename in ./*; do num="$(echo "${filename}" | grep -o '^./[0-9]*')"; mv "${num}.pdb.ostat" "${num}.ostat"; done

Nevertheless, the easiest way is to use (perl) rename.

rename 's/pdb.//' *.pdb.ostat –

See man rename for more info.


Here's a bash script that accomplishes what you want in a less specific way.


# a bash script for changing the extension of files.
echo "what extension would you like to change in this directory?"
read oldext
echo "To what?"
read newext
for f in *.* 
    name=$(echo "$f" | sed 's/\.[^\.]*$//')
    ext=$(echo "$f" | sed 's/^.*\.//')
    echo "source file" "$f" is made up of the base "$name" and ends with "$ext"
    if [ -e "$target" ];
        echo "$target" exists. skipping renameing.
        mv "$f" "$target"
        echo changed "$f" to "$target"


This script does not break on filenames with spaces and allows you to choose the source and target extensions.

It provides examples of a for loop, reading input into variables, using sed to pick a string apart at the last . in the string (filename), how spaces can be handled properly by quoting variables, checking for the existence of a target file and flow control based on the result of a test. It does spam the terminal with unnecessary output that can be eliminated by simple deleting lines 14 and 20 which are really only there to serve as explanation.


Here a script building on prior response. Find all files in the directory with name containing 'Igua' then rename all files including the right extension

for file in *Igua*
  let "i++"
  mv "$file" "Iguazu_$i.$fileext"

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