blocks|key|1429113|text|如果您的sed(1)有一个-i选项，那么可以这样使用它：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1429114|for+i+in+*;+do
++sed+-i+'s/foo/bar/'+$i
done|code-block|syntax|javascript|1429115|如果不是，根据您要使用的语言，有以下几种方式的变体：|1429116|ruby+-i.bak+-pe+'sub(%25r{foo},+'bar')'+*
perl+-pi.bak+-e+'s/foo/bar/'+*|1429117|entityMap^0|4|6|D|2|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

If your <code>sed(1)</code> has a <code>-i</code> option, then use it like this:

<pre><code>for i in *; do
 sed -i 's/foo/bar/' $i
done
</code></pre>

If not, there are several ways variations on the following depending on which language you want to play with:

<pre><code>ruby -i.bak -pe 'sub(%r{foo}, 'bar')' *
perl -pi.bak -e 's/foo/bar/' *
</code></pre>

blocks|key|1644062|text|根据您给出的示例，这似乎就是您想要的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1644063|sed+-i+'s/foo/bar/g'+*|code-block|syntax|javascript|1644064|它不是递归的(它不会下降到子目录中)。对于一个很好的解决方案，在整个树中替换选定的文件，我会使用find：|1644065|find+.+-name+'*.html'+-print+-exec+sed+-i.bak+'s/foo/bar/g'+{}+\;|1644066|*.html是文件必须匹配的表达式，-i之后的.bak创建原始文件的副本，扩展名为.bak+(可以是您喜欢的任何扩展名)，而sed表达式末尾的g告诉sed替换一行上的多个副本(而不仅仅是第一个)。要查找的-print可以方便地显示正在匹配的文件。所有这一切都取决于系统上这些工具的确切版本。|offset|length|style|CODE|1644067|entityMap^0|0|0|0|0|0|6|I|2|N|4|1Z|1|2U|6|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]|$M|Z|N|10|O|P]|$M|11|N|12|O|P]|$M|13|N|14|O|P]|$M|15|N|16|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|R|$]]

This appears to be what you want, based on the example you gave:

<pre><code>sed -i 's/foo/bar/g' *
</code></pre>

It is not recursive (it will not descend into subdirectories). For a nice solution replacing in selected files throughout a tree I would use find:

<pre><code>find . -name '*.html' -print -exec sed -i.bak 's/foo/bar/g' {} \;
</code></pre>

The <code>*.html</code> is the expression that files must match, the <code>.bak</code> after the <code>-i</code> makes a copy of the original file, with a .bak extension (it can be any extension you like) and the <code>g</code> at the end of the sed expression tells sed to replace multiple copies on one line (rather than only the first one). The <code>-print</code> to find is a convenience to show which files were being matched. All this depends on the exact versions of these tools on your system.

blocks|key|1644121|text|我喜欢并使用了上面的解决方案，或者在数千个文件中进行系统范围的搜索和替换：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1644122|find+-name+'*.htm?'+-print+-exec+sed+-i.bak+'s/foo/bar/g'+{}+\;|code-block|syntax|javascript|1644123|我假设是“*.htm？”它搜索并查找类似的.htm和.html文件，而不是.html。|1644124|我将.bak替换为系统范围更广的波浪号(~)，以便更轻松地清理备份文件。|1644125|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I like and used the above solution or a system wide search and replace among thousands of files:

<pre><code>find -name '*.htm?' -print -exec sed -i.bak 's/foo/bar/g' {} \;
</code></pre>

I assume with the '*.htm?' instead of .html it searches and finds .htm and .html files alike.

I replace the .bak with the more system wide used tilde (~) to make clean up of backup files easier.

blocks|key|1429267|text|find+.+-type+f+-print0+%7C+xargs+-0+<sed/perl/ruby+cmd>将一次处理多个包含空格的文件名，每个批处理加载一个解释器。快得多。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1429268|entityMap^0|0|1H|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

<code>find . -type f -print0 | xargs -0 &lt;sed/perl/ruby cmd&gt;</code> will process multiple space contained file names at once loading one interpreter per batch. Much faster.

blocks|key|1429326|text|这在使用grep时有效，不需要使用perl或find。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1429327|grep+-rli+'old-word'+*+%7C+xargs+-i@+sed+-i+'s/old-word/new-word/g'+@|code-block|syntax|javascript|1429328|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|$]]

This works using grep without needing to use perl or find.

<pre><code>grep -rli 'old-word' * | xargs -i@ sed -i 's/old-word/new-word/g' @
</code></pre>

blocks|key|1644336|text|2022答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1644337|https://github.com/BurntSushi/ripgrep/blob/master/FAQ.md#how-can-i-search-and-replace-with-ripgrep|offset|length|1644338|rg+foo+--files-with-matches+%7C+xargs+sed+-i+'s/foo/bar/g'|code-block|syntax|javascript|1644339|将在ripgrep找到foo模式的文件中用'bar‘替换'foo’的所有实例。sed的-i标志表示您正在就地编辑文件，s/+foo+/-i/g表示您正在执行模式foo对bar的替换，并且您正在全局地执行此替换(每个文件中该模式的所有匹配项)。|1644340|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|2Q|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|H|7|Y|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|Z|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|C]]]]

2022 answer. 
<a href="https://github.com/BurntSushi/ripgrep/blob/master/FAQ.md#how-can-i-search-and-replace-with-ripgrep" rel="nofollow noreferrer">https://github.com/BurntSushi/ripgrep/blob/master/FAQ.md#how-can-i-search-and-replace-with-ripgrep</a>
<pre><code>rg foo --files-with-matches | xargs sed -i 's/foo/bar/g'
</code></pre>
will replace all instances of 'foo' with 'bar' in the files in which ripgrep finds the foo pattern. The -i flag to sed indicates that you are editing files in place, and s/foo/bar/g says that you are performing a substitution of the pattern foo for bar, and that you are doing this substitution globally (all occurrences of the pattern in each file).

I'm trying to search and replace a string in all files matched by grep:
<code>grep -n 'foo' *</code> will give me output in the form:
<pre><code>[filename]:[line number]:[text]
</code></pre>
For each file returned by grep, I'd like to modify the file by replacing <code>foo</code> with <code>bar</code>.

Command line: search and replace in all filenames matched by grep

解释器

我正在尝试搜索并替换所有与grep匹配的文件中的字符串：grep -n 'foo' *将以下面的形式给出输出：[filename]:[line number]:[text]对于grep返回的每个文件，我希望通过用bar替换foo来修改该文件。

问命令行:在grep匹配的所有文件名中搜索和替换
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