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使用正则表达式收集所有IP(v4,v6)失败

[英]Collect all IPs (v4,v6) using regex fails

可变source包含我试图在数组ips获取的IP地址,但是出了点问题,我找不到错误...

var ips = source.match(/((^\\s*((([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5])\\.){3}([0-9]|[1-9][0-9]|1[0-9]{2}|2[0-4][0-9]|25[0-5]))\\s*$)|(^\\s*((([0-9A-Fa-f]{1,4}:){7}([0-9A-Fa-f]{1,4}|:))|(([0-9A-Fa-f]{1,4}:){6}(:[0-9A-Fa-f]{1,4}|((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3})|:))|(([0-9A-Fa-f]{1,4}:){5}(((:[0-9A-Fa-f]{1,4}){1,2})|:((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3})|:))|(([0-9A-Fa-f]{1,4}:){4}(((:[0-9A-Fa-f]{1,4}){1,3})|((:[0-9A-Fa-f]{1,4})?:((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3}))|:))|(([0-9A-Fa-f]{1,4}:){3}(((:[0-9A-Fa-f]{1,4}){1,4})|((:[0-9A-Fa-f]{1,4}){0,2}:((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3}))|:))|(([0-9A-Fa-f]{1,4}:){2}(((:[0-9A-Fa-f]{1,4}){1,5})|((:[0-9A-Fa-f]{1,4}){0,3}:((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3}))|:))|(([0-9A-Fa-f]{1,4}:){1}(((:[0-9A-Fa-f]{1,4}){1,6})|((:[0-9A-Fa-f]{1,4}){0,4}:((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3}))|:))|(:(((:[0-9A-Fa-f]{1,4}){1,7})|((:[0-9A-Fa-f]{1,4}){0,5}:((25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)(\\.(25[0-5]|2[0-4]\\d|1\\d\\d|[1-9]?\\d)){3}))|:)))(%.+)?\\s*$))\\b/g);

Since you have not provided a verifiable sample I will provide an working end-to-end demo based on Syzdek's excellent ipv6-regex-test.sh . 由于您尚未提供可验证的样本,因此我将基于Syzdek出色的ipv6-regex-test.sh提供一个有效的端到端演示

The regex expression: 正则表达式:

(([0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}| ([0-9a-fA-F]{1,4}:){1,7}:| ([0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}| ([0-9a-fA-F]{1,4}:){1,5}(:[0-9a-fA-F]{1,4}){1,2}| ([0-9a-fA-F]{1,4}:){1,4}(:[0-9a-fA-F]{1,4}){1,3}| ([0-9a-fA-F]{1,4}:){1,3}(:[0-9a-fA-F]{1,4}){1,4}| ([0-9a-fA-F]{1,4}:){1,2}(:[0-9a-fA-F]{1,4}){1,5}| [0-9a-fA-F]{1,4}:((:[0-9a-fA-F]{1,4}){1,6})| :((:[0-9a-fA-F]{1,4}){1,7}|:)| fe80:(:[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}| ::(ffff(:0{1,4}){0,1}:){0,1} ((25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])\\.){3,3} (25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])| ([0-9a-fA-F]{1,4}:){1,4}: ((25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])\\.){3,3} (25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])) |((25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])\\.){3,3}(25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])

Demo code: 演示代码:

 var ips = new Array();; const regex = /(([0-9a-fA-F]{1,4}:){7,7}[0-9a-fA-F]{1,4}|([0-9a-fA-F]{1,4}:){1,7}:|([0-9a-fA-F]{1,4}:){1,6}:[0-9a-fA-F]{1,4}|([0-9a-fA-F]{1,4}:){1,5}(:[0-9a-fA-F]{1,4}){1,2}|([0-9a-fA-F]{1,4}:){1,4}(:[0-9a-fA-F]{1,4}){1,3}|([0-9a-fA-F]{1,4}:){1,3}(:[0-9a-fA-F]{1,4}){1,4}|([0-9a-fA-F]{1,4}:){1,2}(:[0-9a-fA-F]{1,4}){1,5}|[0-9a-fA-F]{1,4}:((:[0-9a-fA-F]{1,4}){1,6})|:((:[0-9a-fA-F]{1,4}){1,7}|:)|fe80:(:[0-9a-fA-F]{0,4}){0,4}%[0-9a-zA-Z]{1,}|::(ffff(:0{1,4}){0,1}:){0,1}((25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])\\.){3,3}(25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])|([0-9a-fA-F]{1,4}:){1,4}:((25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])\\.){3,3}(25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9]))|((25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])\\.){3,3}(25[0-5]|(2[0-4]|1{0,1}[0-9]){0,1}[0-9])/g; const str = `3ffe:1900:4545:3:200:f8ff:fe21:67cf fe80:0:0:0:200:f8ff:fe21:67cf 192.168.0.1`; let m; while ((m = regex.exec(str)) !== null) { // This is necessary to avoid infinite loops with zero-width matches if (m.index === regex.lastIndex) { regex.lastIndex++; } // capturing group 0: full match console.log(m[0]); ips.push(m[0]); } console.log(ips); 

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