Linux命令行全局搜索和替换 [英] Linux command line global search and replace

问题描述

我试图查找和替换由grep的Linux机器上匹配的所有文件的字符串。我有什么，我想要做的一些作品，但我不能确定如何最好地把它们串在一起。

I'm trying to search and replace a string in all files matched by grep on a linux machine. I've got some pieces of what I want to do, but I'm unsure how best to string them all together.

的grep -n'富'* 会给我输出形式：

[filename]:[line number]:[text]

有关通过的grep返回每个文件，我想取代富与酒吧，结果写回文件。有没有做到这一点的好办法？也许看中管道？

For each file returned by grep, I'd like replace "foo" with "bar" and write the result back to the file. Is there a good way to do that? Maybe a fancy pipeline?

推荐答案

你的意思是搜索和替换通过的grep匹配所有文件的字符串？

Do you mean search and replace a string in all files matched by grep?

perl -p -i -e 's/oldstring/newstring/g' `grep -ril searchpattern *`

修改

由于这似乎是一个相当热门的一个问题想我会更新。

Since this seems to be a fairly popular question thought I'd update.

现在我主要使用 ACK-的grep ，因为它更方便用户使用。所以上面的命令是：

Nowadays I mostly use ack-grep as it's more user-friendly. So the above command would be:

perl -p -i -e 's/old/new/g' `ack -l searchpattern`

要在文件名中处理空格您可以运行：

To handle whitespace in file names you can run:

ack --print0 -l searchpattern | xargs -0 perl -p -i -e 's/old/new/g'

您可以用 ACK-的grep 做多。假设你想将搜索限制HTML文件只：

you can do more with ack-grep. Say you want to restrict the search to HTML files only:

ack --print0 --html -l searchpattern | xargs -0 perl -p -i -e 's/old/new/g'

如果白色空间不是它甚至更短的一个问题：

And if white space is not an issue it's even shorter:

perl -p -i -e 's/old/new/g' `ack -l --html searchpattern`
perl -p -i -e 's/old/new/g' `ack -f --html` # will match all html files

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