正则表达式捕获两个冒号之间所有内容的第一个匹配项

Question

给定这个字符串：

Equinox: *spamReceiveTask: Mar 17 12:34:39.264: #CAPWAP-3-DTLS_CONN_ERR: capwap_ac.c:934 00:3a:9a:30:f5:90: DTLS connection not found forAP 192.168.99.74 (43456), Controller: 192.168.99.2 (5246) send packet

我想匹配 Equinox: 和下一个冒号之间的所有内容。所以我的匹配应该是 *spamReceiveTask

我试过这个： ^.*\bEquinox:([^:]+):

但它匹配为Equinox: *spamReceiveTask:

我也尝试了前后查看选项，但它匹配了字符串中所有分号之间的所有内容。我只想匹配前两个冒号之间的所有字符，排除冒号。

Answer 1

通过使用lookbehind和lookahead断言，请您尝试一下：

(?<=Equinox:)[^:]+(?=:)

Answer 2

您的正则表达式非常好，并且可以正常工作。我想问题出在特定于语言的正则表达式实用程序中，并且您可能以错误的方式使用它们；您想要的输出在捕获组 1 中；

检查下面用javascript编写的sn-p，并在您的模式中使用match()；

let sample = `Equinox: *spamReceiveTask: Mar 17 12:34:39.264: #CAPWAP-3-DTLS_CONN_ERR: capwap_ac.c:934 00:3a:9a:30:f5:90: DTLS connection not found forAP 192.168.99.74 (43456), Controller: 192.168.99.2 (5246) send packet`;

//only the first complete match and its related capturing groups are returned as an array
let pattern = /^.*\bEquinox:([^:]+):/
let matchAndCaptureGroup = sample.match(pattern);
let captureGroup = matchAndCaptureGroup[1];
console.log("without global flag: ", captureGroup);

//------------------------------------------
//with global flag, capturing group is not return
//only the matched strings will return
pattern = /^.*\bEquinox:([^:]+):/g
let matchOnly = sample.match(pattern)[0];
console.log("with global flag: ", matchOnly);