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Replacing strings in files based on certain search criteria is a very common task. How can I

  • replace string foo with bar in all files in the current directory?
  • do the same recursively for sub directories?
  • replace only if the file name matches another string?
  • replace only if the string is found in a certain context?
  • replace if the string is on a certain line number?
  • replace multiple strings with the same replacement
  • replace multiple strings with different replacements
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1  
This is intended to be a canonical Q&A on this subject (see this meta discussion), please feel free to edit my answer below or add your own. –  terdon Feb 1 '14 at 17:08
    
I just want to replace foo with bar in a specific line. all the lines contains foo's.what to do? –  user2433165 Sep 4 '14 at 11:38
    
@user2433165 look at the last bullet point of solution number 3 in my answer below. Just change 4 to the line number you want. –  terdon Sep 4 '14 at 14:05
    
Umm... is it wise to use nonportable tools/options in canonical Q/As? –  mikeserv Sep 4 '14 at 17:30
    
@mikeserv I don't see why not. My answer is basically sed, awk and perl all of which are available on the vast majority of systems. Note that it's a community wiki answer and is so precisely to encourage people to edit and improve it. If you have more portable solutions, please add them. –  terdon Sep 4 '14 at 18:23

2 Answers 2

up vote 71 down vote accepted

1. Replacing all occurrences of one string with another in all files in the current directory:

These are for cases where you know that the directory contains only regular files and that you want to process all non-hidden files. If that is not the case, use the approaches in 2.

All sed solutions in this answer assume GNU sed. If using FreeBSD or OS/X, replace -i with -i ''.

  • Non recursive, files in this directory only:

    sed -i -- 's/foo/bar/g' *
    perl -Ti -pe 's/foo/bar/g' ./* 
    

    (the perl one will fail for file names ending in | or space)).

  • Recursive, regular files (including hidden ones) in this and all subdirectories

    find . -type f -exec sed -i 's/foo/bar/g' {} +
    

    If you are using zsh:

    sed -i -- 's/foo/bar/g' **/*(D.)
    

    (may fail if the list is too big, see zargs to work around).

    If you are using bash, bash having no support for glob qualifiers, you can't check for regular files:

    shopt -s globstar
    shopt -s dotglob
    

    Then:

    sed -i -- 's/foo/bar/g' **/*
    

2. Replace only if the file name matches another string / has a specific extension / is of a certain type etc:

  • Non-recursive, files in this directory only:

    sed -i -- 's/foo/bar/g' *baz*    ## all files whose name contains baz
    sed -i -- 's/foo/bar/g' *.baz    ## files ending in .baz
    
  • Recursive, regular files in this and all subdirectories

    find . -type f -name "*baz*" -exec sed -i 's/foo/bar/g' {} +
    

    If you are using bash:

    shopt -s globstar
    shopt -s dotglob
    

    Then:

    sed -i -- 's/foo/bar/g' **/*baz*
    sed -i -- 's/foo/bar/g' **/*.baz
    

    If you are using zsh:

    sed -i -- 's/foo/bar/g' **/*baz*(D.)
    sed -i -- 's/foo/bar/g' **/*.baz(D.)
    

    The -- serves to tell sed that no more flags will be given in the command line. This is useful to protect against file names starting with -.

  • If a file is of a certain type, for example, executable (see man find for more options):

    find . -type f -executable -exec sed -i 's/foo/bar/g' {} +
    

    zsh:

    sed -i -- 's/foo/bar/g' **/*(D*)
    

3. Replace only if the string is found in a certain context

  • Replace foo with bar only there is a baz later on the same line:

    sed -i ':1;s/foo\(.*baz\)/bar\1/;t1' file
    

    In sed, using \( \) saves whatever is in the parentheses and you can then access it with \1. There are many variations of this theme, to learn more about such regular expressions, see here. We need to repeat the operation for all foo occurrences, which is done with the t conditional branching.

  • Replace foo with bar only if foo is found on the 3d column (field) of the input file (assuming whitespace-separated fields):

    gawk -i inplace 'gsub(/foo/,"baz",$3)' file
    

    (need gawk 4.1.0 or newer).

    For a different field just use $N where N is the number of the field of interest. For a different field separator (: in this example) use:

    gawk -i inplace -F':' 'gsub(/foo/,"baz",$3)' file
    

    Another solution using perl:

    perl -i -ane '$F[2]=~s/foo/baz/g; $" = " "; print "@F\n"' foo 
    

    NOTE: both the awk and perl solutions will print space separated fields even if the input file had tabs. For a different field use $F[N-1] where N is the field umber you want and for a different field separator use (the $"=":" sets the output field separator to :):

    perl -i -F':' -ane '$F[2]=~s/foo/baz/g; $"=":";print "@F"' foo 
    
  • Replace foo with bar only on the 4th line:

    sed -i '4s/foo/bar/g' file
    gawk -i inplace 'NR==4{gsub(/foo/,"baz")};1' file
    perl -i -pe 's/foo/bar/g if $.==4' file
    

4. Multiple replace operations: replace with different strings

You can combine sed commands:

sed -i 's/foo/bar/g; s/baz/zab/g; s/Alice/Joan/g' file

or Perl commands

perl -i -pe 's/foo/bar/g; s/baz/zab/g; s/Alice/Joan/g' file

If you have a large number of patterns, it is easier to save your patterns and their replacements in a sed script file:

#! /usr/bin/sed -i
s/foo/bar/g
s/baz/zab/g

Or, if you have too many pattern pairs for the above to be feasible, you can read pattern pairs from a file (two space separated patterns, $pattern and $replacement, per line):

while read -r pattern replacement; do   
   sed -i "s/$pattern/$replacement/" file
done < patterns.txt

That will be quite slow for long lists of patterns and large data files so you might want to read the patterns and create a sed script from them instead:

sed -f <(awk '{printf "s/%s/%s/g\n", $1, $2}' patterns.txt) -i -- file.txt

Then, run the sed script on your input file(s):

sed -f sedscript.txt inputfile.txt

5. Multiple replace operations: replace multiple patterns with the same string

Replace any of foo, bar or baz with foobar

sed -Ei 's/foo|bar|baz/foobar/g' file

or

perl -i -pe 's/foo|bar|baz/foobar/g' file
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A version with ed should be added because sed -i is not supported on lots of sed version. but I never manage to use ed correctly –  Kiwy Feb 7 '14 at 12:52
    
@kiwy does ed have an -i option? And which sed implementations don't? I know that both GNU and BSD sed do. Also, perl will always have it. Anyway, feel free to add/edit as you wish, that's why I made this community wiki. –  terdon Feb 7 '14 at 13:06
    
that's the nice thing with AIX, when everybody except something simple and heavily spread will work, it will not on this system. -.- anyway, I will add the case where sed does not enable inline edition. –  Kiwy Feb 7 '14 at 13:08
1  
@somethingSomething, the sed command does not rename! It reads every line of every input file and replaces the first occurrence of foo with bar on each line. Of course that will take ages for 13GB of files! –  terdon Nov 2 '14 at 14:13
2  
On Mac OS X it must be sed -i '' 's/foo/bar/' file, as -i expects an extension string for the backup file ('' therefore means no backup file will be created) –  adius Jan 2 at 19:17

"You could also use find and sed, but I find that this little line of perl works nicely.

perl -pi -w -e 's/search/replace/g;' *.php
  • -e means execute the following line of code.
  • -i means edit in-place
  • -w write warnings
  • -p loop over the input file, printing each line after the script is applied to it.

" (Extracted from http://www.liamdelahunty.com/tips/linux_search_and_replace_multiple_files.php)

My best results come from using perl and grep (to ensure that file have the search expression )

perl -pi -w -e 's/search/replace/g;' $( grep -rl 'search' )
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