Unix以递归方式查找和替换？(Unix find and replace recursively?)

Unix以递归方式查找和替换？(Unix find and replace recursively?)

希望这对于这里的某个人来说是一个快速的工具......我需要在unix中递归地查找和替换一个字符串。

通常情况下，我使用：

perl -e "s/term/differenterm/g;" -pi $(find path/to/DIRECTORY -type f)

但是，我需要替换的字符串包含斜杠，我不确定如何逃脱它们？

所以，我需要做的是：

perl -e "s/FIND/REPLACE/g;" -pi $(find path/to/DIRECTORY -type f)

FIND ='/ string / path / term'和REPLACE ='/ string / path / newterm'

Hopefully, this will be a quick one for someone here... I need to find and replace a string recursively in unix.

Normally, I use:

perl -e "s/term/differenterm/g;" -pi $(find path/to/DIRECTORY -type f)

But, the string I need to replace contains slashes, and I'm not sure how to escape them?

So, I need to do:

perl -e "s/FIND/REPLACE/g;" -pi $(find path/to/DIRECTORY -type f)

where FIND = '/string/path/term' and REPLACE = '/string/path/newterm'

最满意答案

您可以使用除/之外的其他字符。 例如：

perl -e "s|FIND|REPLACE|g;" -pi $(find path/to/DIRECTORY -type f)

http://perldoc.perl.org/perlop.html#Regexp-Quote-Like-Operators中的更多信息

You could use other characters besides /. For example:

perl -e "s|FIND|REPLACE|g;" -pi $(find path/to/DIRECTORY -type f)

More info in http://perldoc.perl.org/perlop.html#Regexp-Quote-Like-Operators