[英]PHPStorm Regex Replace with back calls
I want to use PHPStorm to find and replace all instances of: 我想使用PHPStorm查找和替换以下所有实例:
[A-z]['][s]
Example: 例:
Andy's
or David's
Andy's
或David's
to: 至:
Andy\\'s
or David\\'s
Andy\\'s
或David\\'s
I have the regex as above, but I want to know how to use the found character in the regex in the replace. 我有上面的正则表达式,但是我想知道如何在替换中使用在正则表达式中找到的字符。
There are several problems with the regex the way you have it: [Az]['][s]
正则表达式的使用方式有几个问题:
[Az]['][s]
You can't use a shortcut [Az]
to get a range of all upper and lower. 您不能使用快捷方式
[Az]
来获取所有上限和下限的范围。 You need to use [A-Za-z]
. 您需要使用
[A-Za-z]
。 You also don't need the apostrophe and s in brackets: 您也不需要括号中的撇号和:
[A-Za-z]'s
Then, to replace with a matched group, use $ groups: 然后,要用匹配的组替换,请使用$组:
([A-Za-z])'s
, replacing with $01\\\\\\\\'s
([A-Za-z])'s
,替换为$01\\\\\\\\'s
The $
character is used to access the array of regex groups. $
字符用于访问正则表达式组的数组。
The regex groups can be defined with brackets. 正则表达式组可以用方括号定义。
This works: 这有效:
Find: ([Az/0-9])['][s]
查找:
([Az/0-9])['][s]
Replace: $01\\\\\\\\'s
替换:
$01\\\\\\\\'s
(To print one slash, 4 are needed.) (要打印一个斜杠,需要4个。)
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