英文:
Golang Regex not matching the optional part even when it is present in the string
问题
我正在尝试解析一些命令的输出，我想检查命令是否有错误，所以我查找字符串**Apply Error**。如果没有错误，之前提到的字符串就不存在。我的正则表达式匹配模式是(?s)Ran Apply.+?(\*\*Apply Error\*\*)?。我是在Golang中进行这个操作的。
即使输出中存在字符串** Apply Error**，这个正则表达式也不起作用，如果我删除正则表达式中的最后一个?，它就能按预期工作。它捕获了两个组，第二个组是可选的部分。
示例：
情况1：当字符串**Apply Error**存在时
Ran Apply for dir: `some random path/something/something`
`**Apply Error**`
输出：匹配**Apply Error**并将其捕获为一个组
情况2：当**Apply Error**不存在时
Ran Apply for dir: `some random path/something/something`
some random text
输出：按原样匹配所有内容
演示：https://regex101.com/r/Um3CNc/1
在演示中，即使存在** Apply Error**，它也无法匹配。
如何匹配可选部分？
英文:
I am trying to parse some output from a command, I want to check if the command had an error in it, so I look for the string **Apply Error**. If there is no error present the previously mentioned string is absent. My regex to match that is (?s)Ran Apply.+?(\*\*Apply Error\*\*)?. I am doing this in Golang
This regex does not work even when the string ** Apply Error** is present in the output, if I remove the last ? in my regex it works as expected. It captures 2 groups with the second being the optional part.
Examples:
Case - 1 when the string **Apply Error** is present
Ran Apply for dir: `some random path/something/something`
`**Apply Error**`
Output: Match **Apply Error** and capture it as a group
Case - 2
When **Apply Error** is not present
Ran Apply for dir: `some random path/something/something`
some random text
Output: Match everything as is
Demo: https://regex101.com/r/Um3CNc/1
In the demo even though ** Apply Error** is present it does not match it
How can I match the optional part?
答案1
得分: 2
你可以使用以下正则表达式进行匹配：
(?s)Ran Apply\b(?:.+?(\*\*Apply Error\*\*)|.*)
解释：
(?s)内联修饰符，使点号匹配换行符Ran Apply\b匹配字面上的 "Ran Apply" 后面跟着一个单词边界(?:非捕获组.+?(\*\*Apply Error\*\*)匹配1个或多个字符，尽可能少地匹配，并在第1组中捕获**Apply Error**|或者.*匹配剩余的文本
)结束非捕获组
查看带有 "Apply Error" 的正则表达式演示链接，以及不带 "Apply Error" 的正则表达式演示链接。
英文:
You can use
(?s)Ran Apply\b(?:.+?(\*\*Apply Error\*\*)|.*)
Explanation
(?s)Inline modifier to have the dot match a newlineRan Apply\bMatch literally followed by a word boundary(?:Non capture group.+?(\*\*Apply Error\*\*)Match 1+ chars, as few as possible and capture**Apply Error**in group 1|Or.*Match the rest of the text
)Close non capture group
See a regex demo with Apply Error present and without Apply Error present.
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