JS正则提取URL中斜杠之间的值？

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大佬们，如何用js正则匹配，得出每个“/”中间的值

let pattern = "{note}/:pakid:/corrinsdetail/{corrInsAleid}/{corrInsPakId}/{corrInsInsta}/{corrInsVernr}/{verFrom}/{verTo}/{name}/:tab:";
let url = "0001668882/41/corrinsdetail/0001023444/41/0020751259/0005/756/756/SAP_BASIS/ValidityDetails";

输入上述数据，期望得到结果

{
  note: "0001668882",
  pakid: "41",
  corrInsAleid: "0001023444",
  corrInsPakId: "41",
  corrInsInsta: "0020751259",
  corrInsVernr: "0005",
  verFrom: "756",
  verTo: "756",
  name: "SAP_BASIS",
  tab: "ValidityDetails"
}
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1个回答
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test
2024-06-19

代码代码显得有点长，为了兼容异常情况，

  1. 检验url是否符合pattern，
  2. url如果含有参数 ?zz=23423 或者 #abc，需要清洗

  3. url中如果含有-也要兼容，不仅仅是_和数字、字母

    let pattern = "{note}/:pakid:/corrinsdetail/{corrInsAleid}/{corrInsPakId}/{corrInsInsta}/{corrInsVernr}/{verFrom}/{verTo}/{name}/:tab:";
let url = "0001668882/41/corrinsdetail/0001023444/41/0020751259/0005/756/756/SAP_BASIS/ValidityDetails";


function extractInfo(pattern, url) {
 const urlRegex = new RegExp(pattern.replace(/(\{|:)[\w-]+(\}|:)/g, "\[\\w-]+").replace(/\//g, "\/"));
 if(!urlRegex.test(url)) {
     throw new Error("输入的Url不符合Pattern");
 }
 let keys = pattern.split("/");
 let values = url.replace(/(\?|#).*/, "").split("/");
 let result = {};
 keys.forEach((v,i)=>{
     if(/(\{|:)\w+(\}|:)/.test(v)) {
         result[v.replace(/(\{|:|\})/g,"")] = values[i];
     }
 })
 return result;
}

console.log(extractInfo(pattern, url));

输出结果answer image

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