1

I have a file (ordered_names) which is in the format

pub 000.html
pub.19 001.html

for about 300 lines, And I can't find a way to feed this to the mv command.

I have read Provide strings stored in a file as a list of arguments to a command?, but I could not get what I came for.

Here are some of the attempts I made :

for line in "$(cat ../ordered_files.reversed)"; do mv  $(echo "$line"); done
for line in "$(cat ../ordered_files.reversed)"; do echo mv $(echo $line | cut -d' ' -f 1) $(echo $line | cut -d' ' -f 2) ; done
Pierre-Antoine Guillaume
  • 331

1 Answers1

4

Try:

while read -r file1 file2; do mv -n -- "$file1" "$file2"; done <inputfile

This assumes that the file names on each line in the input file are space-separated. This, of course, only works if the file names themselves do not contain spaces. If they do, then you need a different input format.

How it works

  • while read -r file1 file2; do

    This starts a while loop. The loop continues as long as input is available. For each line of input, two parameters are read: file1 and file2.

  • mv -n -- "$file1" "$file2"

    This moves file1 to file2.

    The option -n protects you from overwriting any file at the destination. Of course, if it is your intention to overwrite files, remove this option.

    The string -- signals the end of the options. This protects you from problems should any of the file names start with a -.

  • done

    This signals the end of the while loop.

  • <inputfile

    This tells the while loop to gets its input from a file called inputfile.

John1024
  • 74,655
  • I always recommend using mv -n or mv -i for bulk moves like this, to avoid accidental overwriting (silent & irreversible) of files due to name conflicts. In this case, mv -i would try to read overwrite confirmations from the file list, which wouldn't work well at all, so mv -n is the way to go. – Gordon Davisson Dec 16 '17 at 07:24
  • @GordonDavisson Excellent point. I just added -n added to the answer. – John1024 Dec 16 '17 at 07:34