基于THIS或THAT的正则表达式 [英] Regex based on THIS or THAT

问题描述

我正在尝试解析以下内容：

I am trying to parse the below:

"#SenderCompID=something\n" +
"TargetCompID=something1"

放入数组：

{"#SenderCompID=something", "TargetCompId", "something1"}

使用：

String regex = "(?m)" + "(" +     
    "(#.*) |" +                //single line of (?m)((#.*)|([^=]+=(.+))
    "([^=]+)=(.+) + ")";
String toMatch = "#SenderCompID=something\n" +
    "TargetCompID=something1";

正在输出：

#SenderCompID=something
null
#SenderCompID
something
                       //why is there any empty line here?
TargetCompID=something1
null
                       //why is there an empty line here?
TargetCompID
something1

我知道我在做什么这里错了第一组是返回整行，如果该行以＃开头，则第二组返回（＃。*），否则返回null，否则，第三组返回（[^ =] + =（。+）。 |这就是我想要做的。我想根据第二组的条件来解析它

I understand what I'm doing wrong here. The 1st group is returning the entire line, the 2nd group is returning (#.*) if the line starts with # and null otherwise, the 3rd group is returning ([^=]+=(.+). The | is what I'm trying to do. I want to parse it based on EITHER the conditions for the 2nd group

(#.*)

或用于第三组

([^=]+)=(.+).

如何？

编辑：错误地编写了示例代码

miswrote example code

推荐答案

您可以使用此正则表达式获取所有3个组：

You can use this regex to get all 3 groups:

(?m)^(#.*)|^([^=]+)=(.*)

RegEx演示

正则表达式分解：

• （？m）：启用 MULTILINE 模式


• ^（＃。*）：匹配＃1组中以＃开头的整行


• | ：或


• ^（[^ =] +）= ：匹配到 = 并捕获到＃2组中，然后捕获 =


• （。*）：匹配＃3组中的其余行

• (?m): Enable MULTILINE mode
• ^(#.*): match a full line starting with # in group #1
• |: OR
• ^([^=]+)=: Match till = and capture in group #2 followed by =
• (.*): Match rest of line in group #3

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