mjhjmtu

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

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.