Bash, renaming files won't work

mago · 04-01-2010, 01:11 AM

Hi, I'm trying to rename a lot of files getting rid of the space on the names. For that purpose I wrote this very simple bash script, but for some reason is not working.

Code:

for i in "$(ls)"
do
j=$(echo "$i" | sed 's/\ /_/g')
mv "$i" "$j"
done

But what I get in return for each line is just one long file name with all the file names concatenated.
I've tried with echo -e "$i" as well with no results.

This has to be something really simple that I'm missing but I just can't see it.

Thanks in advanced for any pointers.

GrapefruiTgirl · 04-01-2010, 01:21 AM

Hi there!

You're working along the right idea, but using `ls` for this is not a great idea. Go to LQ's Search Engine and enter:

Code:

rename filename spaces

into the keyword search field; select "Search Entire Posts" from the dropdown gadget below that text field, and click "Search Now" -- you will find many examples of how to do just what you're doing here. Adjust your search keywords as necessary, but I think you'll find the answer on one of the pages returned.

Sasha

smeezekitty · 04-01-2010, 01:26 AM

Try:

Code:

for i in "*.*"
do
j=$(echo "$i" | sed 's/\ /_/g')
mv "$i" "$j"
done

colucix · 04-01-2010, 02:39 AM

Actually the reason why you get all file names concatenated together is that double quotes prevent field splitting (that is a string embedded in double quotes is seen as a single string, despite the presence of blank spaces). This means the loop is cycled only once and the loop variable get the value:

Code:

file one file two file three

Anyway, since you have file names with blank spaces, you have to find a way to prevent file names splitting in two or more pieces. One of them could be:

Code:

while read file
do
  echo "$file"
done < <(ls *\ *)

Here the "read" statement assigns every line of the input to the variable "file". Just use double quotes around "$file" inside the code block and the trick is done.

catkin · 04-01-2010, 02:40 AM

Thanks to tuxdev in this LQ post a robust solution (not tested) is

Code:

while IFS="" read -r -d "" file ; do
   /usr/bin/mv "$file" "${file// }" 
done < <(find . -maxdepth 1 -type f -name '* *' -print0)

catkin · 04-01-2010, 02:44 AM

Anither solution (slightly tested!) is to use plain file name expansion

Code:

for file in *
do
   [[ $file != ${file// } ]] && /usr/bin/mv "$file" "${file// }"
done

colucix · 04-01-2010, 03:08 AM

An even more simple solution is mass renaming using the rename command. However, there are two versions of rename: one coming from util-linux for which you can simply do:

Code:

rename ' ' '_' *\ *

and another (coming from perl) which uses perl expressions, for which the syntax would be:

Code:

rename 's/ /_/g' *\ *

Looking at the manual page of rename, you can easily find out which version is installed on your system.

colucix · 04-01-2010, 03:24 AM

Quote:

Originally Posted by catkin

Anither solution (slightly tested!) is to use plain file name expansion

Code:

for file in *
do
   [[ $file != ${file// } ]] && /usr/bin/mv "$file" "${file// }"
done

Hi catkin. Your solution works, except for the substitution of blank spaces with underscores. Hence, the substring replacement should be

Code:

"${file// /_}"

Surprisingly it works even if I do (but maybe this is not robust)

Code:

for file in *\ *
do
  mv "$file" "${file// /_}"
done

which is almost the same as yours, except the file list contains only names with spaces. But shouldn't it split each name in the file list into pieces?!

catkin · 04-01-2010, 04:28 AM

Quote:

Originally Posted by colucix

But shouldn't it split each name in the file list into pieces?!

The closest to an explanation of how it does behave is here where in step 4 Bash "# Performs the various shell expansions (see Shell Expansions), breaking the expanded tokens into lists of filenames (see Filename Expansion) and commands and arguments". Filename Expansion comes after breaking the command up into IFS-character-separated words and a single glob can be expanded into several words, each of which may contain IFS-characters ...

EDIT:

The above is not accurate. Where I wrote "after breaking the command up into IFS-character-separated words" it should have been "after breaking the command up into metacharacter-character-separated words where a metacharacter is one of:

|
&
;
(
)
<
>
space
tab

colucix · 04-01-2010, 05:36 AM

Quote:

Originally Posted by catkin

Filename Expansion comes after breaking the command up into IFS-character-separated words and a single glob can be expanded into several words, each of which may contain IFS-characters ...

Yesss! Quite clear, now. Thank you!

ghostdog74 · 04-01-2010, 07:59 PM

you can use Python

Code:

import os
os.chdir("/mypath")
for files in os.listdir("."):
    if os.path.isfile(files) and " " in files:
         newfile=files.replace(" ","_")
         os.rename(files,newfile)

Olion · 04-02-2010, 01:10 PM

Hi,
You can also use shell

Code:

$ ls -1
file one.txt
file three.txt
file two.txt
file1.txt
file2.txt
file3.txt
$ find . -name "*\ *" > x
$ cp x y
$ sed -i 's/.*/"&"/' x
$ tr " " "_" < y > z
$ paste x z | sed 's/^/mv /' > s; sh s
$ ls -1
file_one.txt
file_three.txt
file_two.txt
file1.txt
file2.txt
file3.txt
s
x
y
z
$

Greeting
Oleg