I am attempting to match the following string, it always
begins with Removing:
and ends with --- For User:
我试图匹配以下字符串,它始终以删除开头:并以---用户结束:
I tried using the following regex:
我尝试使用以下正则表达式:
\bRemoving: \w+\ --- For User: \b
Removing: Random_Group_Here --- For User: - Should Match This
删除:Random_Group_Here ---对于用户: - 应匹配此项
I'm trying to assign this regex to a powershell variable that I can use for a -replace string.
我正在尝试将此正则表达式赋给一个powershell变量,我可以将其用于-replace字符串。
$regex = "\[\bRemoving: \w+\ --- For User: \b\]"
1 个解决方案
#1
2
The last \b word boundary only matches if there is a word char (i.e. mostly a letter, digit, _) after the space at the end of the pattern. Since there is no such a char, the last word boundary prevents from matching.
如果在模式结尾处的空格之后存在单词char(即,主要是字母,数字,_),则最后的\ b字边界仅匹配。由于没有这样的字符,最后一个字边界阻止匹配。
Either remove the last \b, or - if you want to only match that space when not followed with a word char - with (?!\w) negative lookahead that will check just that.
要么删除最后一个\ b,要么 - 如果你只想匹配那个没有跟着单词char的空格 - 用(?!\ w)否定前瞻来检查那个。
You may also replace literal spaces with \s* (0+ whitespaces) / \s+ (1+ whitespaces) patterns to match any kind and any amount of whitespace between specific chars.
您还可以使用\ s *(0+空格)/ \ s +(1+空格)模式替换文字空间,以匹配特定字符之间的任何种类和任何数量的空白。
Here is a variation:
这是一个变化:
\bRemoving:\s*\w+\s*---\s*For\s+User:\s*
See the regex demo.
请参阅正则表达式演示。
#1
2
The last \b word boundary only matches if there is a word char (i.e. mostly a letter, digit, _) after the space at the end of the pattern. Since there is no such a char, the last word boundary prevents from matching.
如果在模式结尾处的空格之后存在单词char(即,主要是字母,数字,_),则最后的\ b字边界仅匹配。由于没有这样的字符,最后一个字边界阻止匹配。
Either remove the last \b, or - if you want to only match that space when not followed with a word char - with (?!\w) negative lookahead that will check just that.
要么删除最后一个\ b,要么 - 如果你只想匹配那个没有跟着单词char的空格 - 用(?!\ w)否定前瞻来检查那个。
You may also replace literal spaces with \s* (0+ whitespaces) / \s+ (1+ whitespaces) patterns to match any kind and any amount of whitespace between specific chars.
您还可以使用\ s *(0+空格)/ \ s +(1+空格)模式替换文字空间,以匹配特定字符之间的任何种类和任何数量的空白。
Here is a variation:
这是一个变化:
\bRemoving:\s*\w+\s*---\s*For\s+User:\s*
See the regex demo.
请参阅正则表达式演示。