正则表达式在最后一个冒号后找到一个字符串 [英] Regex to find a string after the last colon

问题描述

以下是一些示例输入：

<210>   DW_AT_name        : (indirect string, offset: 0x55): double
 <ae>   DW_AT_name        : (indirect string, offset: 0x24): long int
 <b5>   DW_AT_name        : int

我想提取代表实际类型的字符串。所以我的输出是：

I want to extract the string that represents the actual type. So my output would be:

double
long int
int

这是我到目前为止的正则表达式（双重逃脱，因为它是用Java编写的）：

Here is the regex I have so far (double escaped because it's in Java):

.*DW_AT_name.*:\\s*([^:&&\\S]*)\\s*

适用于 int ，但不起作用对于另外两个。我认为最好的解决方案是基本上说在最后一次结肠后得到所有东西，但我不确定如何。请注意，它还必须包含 DW_AT_NAME 内容。

It works for the int, but it doesn't work for the other two. I think the best solution is to basically say 'get everything after the last colon' but I'm not sure how. Note that it must also include the DW_AT_NAME stuff.

推荐答案

您需要没有正则表达式：

You need no regex for this:

String yourString = "<210>   DW_AT_name        : (indirect string, offset: 0x55): double";
String result;
if (yourString.contains("DW_AT_name")) {
    int lastIndex = yourString.lastIndexOf(":");
    result = yourString.substring(lastIndex + 1).trim();
} else {
    result = "ERROR"; // or handle this however you want
}
System.out.println(result);

只需找到最后一个：并采取一切措施之后。然后修剪它以删除前导和尾随空格。

Simply find the last : and take everything after that. Then trim it to remove leading and trailing whitespace.

我编辑了我的问题，它还需要检查DW_AT_name。字符串拆分不会[原文如此]。

I edited my question, it needs to also check for DW_AT_name. String split wont [sic] do that.

只需使用包含，然后。 （编辑答案）

Just use contains, then. (edited answer)

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