There are a variety of ways to replace one string of text with another across many files. Here are a few ways:

using sed and find:

sed 's/oldstring/newstring/' "$1" > "$1".new && find -iname "*.new" | sed 's/.new//' | sh

using grep and sed:

grep -rl oldstring . | xargs sed -i -e 's/oldstring/newstring/'

using grep and perl:

grep -rl oldstring . | xargs perl -pi~ -e 's/oldstring/newstring/'

Please offer your own suggestions.

  • You should make this a community wiki
    – squillman
    Oct 2, 2009 at 22:35
  • Usually if you are going to ask a question and answer it yourself you should but your answer(s), as an answer. So people can vote on the question and answer separately.
    – Zoredache
    Oct 2, 2009 at 22:38
  • @squillman: done, sorry i forgot to make the wiki community flag on it Oct 2, 2009 at 23:01

6 Answers 6


I'd use Python for this. Put all this code into a file called mass_replace and "chmod +x mass_replace":


import os
import re
import sys

def file_replace(fname, s_before, s_after):
    out_fname = fname + ".tmp"
    out = open(out_fname, "w")
    for line in open(fname):
        out.write(re.sub(s_before, s_after, line))
    os.rename(out_fname, fname)

def mass_replace(dir_name, s_before, s_after):
    for dirpath, dirnames, filenames in os.walk(dir_name):
        for fname in filenames:
            f = fname.lower()
            # example: limit replace to .txt, .c, and .h files
            if f.endswith(".txt") or f.endswith(".c") or f.endswith(".h"):
                f = os.path.join(dirpath, fname)
                file_replace(f, s_before, s_after)

if len(sys.argv) != 4:
    u = "Usage: mass_replace <dir_name> <string_before> <string_after>\n"

mass_replace(sys.argv[1], sys.argv[2], sys.argv[3])

For a single search and replace of one string in one type of file, the solution with find and sed isn't bad. But if you want to do a lot of processing in one pass, you can edit this program to extend it, and it will be easy (and likely to be correct the first time).


Using the GNU find, xargs and sed like this:

 find -name '*.txt' -o -name '*.html' -print0 | xargs -0 -P 1 -n 10 sed --in-place 's/oldstring/newstring/g'

Adjust the -P and -n parameters as you like. The /g is needed so that every occurrence in a line gets replaced, not just the first one (g stands for global if I remember correctly). You can also pass a value to --in-place to make a backup.


I like perl's in-place filtering recipe.

   perl -pi.bak -e 's/from/to/' file1 file2 ...

In context...

% echo -e 'foo\ngoo\nboo' >test
% perl -pi.bak -e 's/goo/ber/' test
% diff -u test.bak test
--- test.bak    2010-01-06 05:43:53.072335686 -0800
+++ test    2010-01-06 05:44:03.751585440 -0800
@@ -1,3 +1,3 @@

here is the trimmed quick-reference on the perl incantation used...

% perl --help
Usage: perl [switches] [--] [programfile] [arguments]
  -e program        one line of program (several -e's allowed, omit programfile)
  -i[extension]     edit <> files in place (makes backup if extension supplied)
  -n                assume "while (<>) { ... }" loop around program
  -p                assume loop like -n but print line also, like sed

Assuming the list of files isn't a mile long, you don't need to use xargs, as sed can handle multiple files on the command line:

sed -i -e 's/oldstring/newstring/' `grep -rl oldstring .`

Be careful if you replace URLs with "/" character.

An example of how to do it:

sed -i "s%http://domain.com%http://www.domain.com/folder/%g" "test.txt"

Extracted from: http://www.sysadmit.com/2015/07/linux-reemplazar-texto-en-archivos-con-sed.html


Thanks for some great answers everyone! This was super-helpful.

Since I didn't have hundreds of files to replace lines in, I used a do loop, like this:

   for R in 1 2 3 4 5; do sed -i -e 's/oldstring/newstring/' file$R; done

Hope that helps!

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