blocks|key|358780|text|另一种选择是使用find，然后通过sed传递它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|358781|find+/path/to/files+-type+f+-exec+sed+-i+'s/oldstring/new+string/g'+{}+\;|code-block|syntax|javascript|358782|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another option is to use find and then pass it through sed.

<pre><code>find /path/to/files -type f -exec sed -i 's/oldstring/new string/g' {} \;
</code></pre>

blocks|key|140829|text|我知道答案了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|140830|grep+-rl+matchstring+somedir/+%7C+xargs+sed+-i+'s/string1/string2/g'|code-block|syntax|javascript|140831|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I got the answer.

<pre><code>grep -rl matchstring somedir/ | xargs sed -i 's/string1/string2/g'
</code></pre>

blocks|key|140855|text|这最适合我在OS+X上使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|140856|grep+-r+-l+'searchtext'+.+%7C+sort+%7C+uniq+%7C+xargs+perl+-e+"s/matchtext/replacetext/"+-pi|code-block|syntax|javascript|140857|来源：http://www.praj.com.au/post/23691181208/grep-replace-text-string-in-files|offset|length|140858|entityMap|0|LINK|mutability|MUTABLE|url|http://www.praj.com.au/post/23691181208/grep-replace-text-string-in-files^0|0|0|3|21|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

This works best for me on OS X:

<pre><code>grep -r -l 'searchtext' . | sort | uniq | xargs perl -e "s/matchtext/replacetext/" -pi
</code></pre>

Source: <a href="http://www.praj.com.au/post/23691181208/grep-replace-text-string-in-files">http://www.praj.com.au/post/23691181208/grep-replace-text-string-in-files</a>

blocks|key|140823|text|通常不是使用grep，而是使用sed+-i+'s/string_to_find/another_string/g'或perl+-i.bak+-pe+'s/string_to_find/another_string/g'。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|140824|entityMap^0|F|16|1M|1F|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Usually not with grep, but rather with <code>sed -i 's/string_to_find/another_string/g'</code> or <code>perl -i.bak -pe 's/string_to_find/another_string/g'</code>.

blocks|key|140908|text|在git+repo中使用find和sed时要非常小心！如果不排除二进制文件，可能会出现以下错误：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|140909|error:+bad+index+file+sha1+signature+
fatal:+index+file+corrupt|code-block|syntax|javascript|140910|要解决此错误，您需要通过将new_string替换为old_string来恢复sed。这将恢复您替换的字符串，因此您将返回到问题的开始。|140911|搜索字符串并替换它的正确方法是跳过find，改为使用grep，以便忽略二进制文件：|140912|sed+-ri+-e+"s/old_string/new_string/g"+$(grep+-Elr+--binary-files=without-match+"old_string"+"/files_dir")|140913|@hobs致谢|140914|entityMap^0|C|4|H|3|0|0|D|A|Q|A|13|3|0|H|4|Q|4|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|1D|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1E|8|@]|D|@]|E|$]]]|T|$]]

Be very careful when using <code>find</code> and <code>sed</code> in a git repo! If you don't exclude the binary files you can end up with this error:

<pre><code>error: bad index file sha1 signature 
fatal: index file corrupt
</code></pre>

To solve this error you need to revert the <code>sed</code> by replacing your <code>new_string</code> with your <code>old_string</code>. This will revert your replaced strings, so you will be back to the beginning of the problem.

The correct way to search for a string and replace it is to skip <code>find</code> and use <code>grep</code> instead in order to ignore the binary files:

<pre><code>sed -ri -e "s/old_string/new_string/g" $(grep -Elr --binary-files=without-match "old_string" "/files_dir")
</code></pre>

Credits for @hobs

blocks|key|140950|text|以下是我会做的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|140951|find+/path/to/dir+-type+f+-iname+"*filename*"+-print0+%7C+xargs+-0+sed+-i+'/searchstring/s/old/new/g'|code-block|syntax|javascript|140952|这将在/path/to/dir下查找文件名中包含filename的所有文件，然后对于找到的每个文件，搜索包含searchstring的行，并将old替换为new。|offset|length|style|CODE|140953|不过，如果您不想查找文件名中包含filename字符串的特定文件，则只需执行以下操作：|140954|find+/path/to/dir+-type+f+-print0+%7C+xargs+-0+sed+-i+'/searchstring/s/old/new/g'|140955|这将做与上面相同的事情，但对在/path/to/dir下找到的所有文件。|140956|entityMap^0|0|0|3|C|O|8|1I|C|1Z|3|25|3|0|G|8|0|0|F|C|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]|$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|17|8|@$I|18|J|19|K|L]]|9|@]|A|$]]|$1|O|3|P|5|D|7|1A|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1B|8|@$I|1C|J|1D|K|L]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|T|$]]

Here is what I would do:

<pre><code>find /path/to/dir -type f -iname "*filename*" -print0 | xargs -0 sed -i '/searchstring/s/old/new/g'
</code></pre>

this will look for all files containing <code>filename</code> in the file's name under the <code>/path/to/dir</code>, than for every file found, search for the line with <code>searchstring</code> and replace <code>old</code> with <code>new</code>.

Though if you want to omit looking for a specific file with a <code>filename</code> string in the file's name, than simply do:

<pre><code>find /path/to/dir -type f -print0 | xargs -0 sed -i '/searchstring/s/old/new/g'
</code></pre>

This will do the same thing above, but to all files found under <code>/path/to/dir</code>.

blocks|key|358826|text|另一种选择是只将perl与globstar一起使用。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|358827|在您的.bashrc+(或任何地方)中启用shopt+-s+globstar允许**+glob模式递归地匹配所有子目录和文件。|offset|length|style|CODE|358828|因此，使用perl+-pXe+'s/SEARCH/REPLACE/g'+-i+**将递归地将SEARCH替换为REPLACE。|358829|perl标志告诉-X“禁用所有警告”--这意味着它不会抱怨目录。|358830|如果您希望在扩展名为.ext的所有子文件中将SEARCH替换为REPLACE，则globstar还允许您执行诸如sed+-i+'s/SEARCH/REPLACE/g'+**/*.ext之类的操作。|358831|entityMap^0|0|3|7|L|H|14|2|0|5|10|1A|6|1J|7|0|8|2|0|A|4|M|6|V|7|1K|10|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@$D|R|E|S|F|G]|$D|T|E|U|F|G]|$D|V|E|W|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|X|8|@$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]]|9|@]|A|$]]|$1|J|3|K|5|6|7|14|8|@$D|15|E|16|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|17|8|@$D|18|E|19|F|G]|$D|1A|E|1B|F|G]|$D|1C|E|1D|F|G]|$D|1E|E|1F|F|G]]|9|@]|A|$]]|$1|N|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]]|O|$]]

Another option would be to just use perl with globstar.

Enabling <code>shopt -s globstar</code> in your <code>.bashrc</code> (or wherever) allows the <code>**</code> glob pattern to match all sub-directories and files recursively.

Thus using <code>perl -pXe 's/SEARCH/REPLACE/g' -i **</code> will recursively
replace <code>SEARCH</code> with <code>REPLACE</code>.

The <code>-X</code> flag tells perl to "disable all warnings" - which means that
it won't complain about directories.

The globstar also allows you to do things like <code>sed -i 's/SEARCH/REPLACE/g' **/*.ext</code> if you wanted to replace <code>SEARCH</code> with <code>REPLACE</code> in all child files with the extension <code>.ext</code>.

I need to recursively search for a specified string within all files and subdirectories within a directory and replace this string with another string.

I know that the command to find it might look like this:

<pre><code>grep 'string_to_find' -r ./*
</code></pre>

But how can I replace every instance of <code>string_to_find</code> with another string?

How to grep and replace

我需要递归地在一个目录内的所有文件和子目录中搜索指定的字符串，并用另一个字符串替换这个字符串。我知道查找它的命令可能如下所示：grep 'string_to_find' -r ./*但是如何将string_to_find的每个实例替换为另一个字符串呢？

问如何进行grep和替换
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