How can I replace a string in a file(s)?
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 ''
. Also note that the use of the -i
switch with any version of sed
has certain filesystem security implications and is inadvisable in any script which you plan to distribute in any way.
Non recursive, files in this directory only:
sed -i -- 's/foo/bar/g' * perl -i -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).Bash can't check directly for regular files, a loop is needed (braces avoid setting the options globally):
( shopt -s globstar dotglob; for file in **; do if [[ -f $file ]] && [[ -w $file ]]; then sed -i -- 's/foo/bar/g' "$file" fi done )
The files are selected when they are actual files (-f) and they are writable (-w).
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 (braces avoid setting the options globally):
( shopt -s globstar dotglob 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
withbar
only if there is abaz
later on the same line:sed -i 's/foo\(.*baz\)/bar\1/' 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.
Replace
foo
withbar
only iffoo
is found on the 3d column (field) of the input file (assuming whitespace-separated fields):gawk -i inplace '{gsub(/foo/,"baz",$3); print}' file
(needs gawk
4.1.0 or newer).
For a different field just use
$N
whereN
is the number of the field of interest. For a different field separator (:
in this example) use:gawk -i inplace -F':' '{gsub(/foo/,"baz",$3);print}' file
Another solution using perl
:
perl -i -ane '$F[2]=~s/foo/baz/g; $" = " "; print "@F\n"' foo
NOTE: both the awk
and perl
solutions will affect spacing in the file (remove the leading and trailing blanks, and convert sequences of blanks to one space character in those lines that match). For a different field, use $F[N-1]
where N
is the field number 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
withbar
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
Be aware that order matters (sed 's/foo/bar/g; s/bar/baz/g'
will substitute foo
with baz
).
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 -f 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. The following assumes a <<!>space<!>> delimiter separates a list of MATCH<<!>space<!>>REPLACE pairs occurring one-per-line in the filepatterns.txt
:sed 's| *\([^ ]*\) *\([^ ]*\).*|s/\1/\2/g|' <patterns.txt | sed -f- ./editfile >outfile
The above format is largely arbitrary and, for example, doesn't allow for a <<!>space<!>> in either of MATCH or REPLACE. The method is very general though: basically, if you can create an output stream which looks like a sed
script, then you can source that stream as a sed
script by specifying sed
's script file as -
stdin.
You can combine and concatenate multiple scripts in similar fashion:
SOME_PIPELINE | sed -e'#some expression script' \ -f./script_file -f- \ -e'#more inline expressions' \ ./actual_edit_file >./outfile
A POSIX sed
will concatenate all scripts into one in the order they appear on the command-line. None of these need end in a \n
ewline.
grep
can work the same way:sed -e'#generate a pattern list' <in | grep -f- ./grepped_file
When working with fixed-strings as patterns, it is good practice to escape regular expression metacharacters. You can do this rather easily:
sed 's/[]$&^*\./[]/\\&/g s| *\([^ ]*\) *\([^ ]*\).*|s/\1/\2/g| ' <patterns.txt | sed -f- ./editfile >outfile
5. Multiple replace operations: replace multiple patterns with the same string
Replace any of
foo
,bar
orbaz
withfoobar
sed -Ei 's/foo|bar|baz/foobar/g' file
or
perl -i -pe 's/foo|bar|baz/foobar/g' file
A good replacement Linux tool is rpl, that was originally written for the Debian project, so it is available with apt-get install rpl
in any Debian derived distro, and may be for others, but otherwise you can download the tar.gz
file from SourceForge.
Simplest example of use:
$ rpl old_string new_string test.txt
Note that if the string contains spaces it should be enclosed in quotation marks. By default rpl
takes care of capital letters but not of complete words, but you can change these defaults with options -i
(ignore case) and -w
(whole words). You can also specify multiple files:
$ rpl -i -w "old string" "new string" test.txt test2.txt
Or even specify the extensions (-x
) to search or even search recursively (-R
) in the directory:
$ rpl -x .html -x .txt -R old_string new_string test*
You can also search/replace in interactive mode with -p
(prompt) option:
The output shows the numbers of files/string replaced and the type of search (case in/sensitive, whole/partial words), but it can be silent with the -q
(quiet mode) option, or even more verbose, listing line numbers that contain matches of each file and directory with -v
(verbose mode) option.
Other options that are worth remembering are -e
(honor escapes) that allow regular expressions
, so you can search also tabs (\t
), new lines (\n
),etc. You can use -f
to force permissions (of course, only when the user has write permissions) and -d
to preserve the modification times`).
Finally, if you are unsure what exactly will happen, use the -s
(simulate mode).
How to do a search and replace over multiple files suggests:
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.
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' )