原生JS获取URL链接参数的几种常见方法

<script>
  // 利用正则表达式
  let url = "http://www.baidu.com?name=elephant&age=25&sex=male&num=100"
  // // 返回参数对象
  function queryURLParams(url) {
    let pattern = /(\w+)=(\w+)/ig; //定义正则表达式
    let parames = {}; // 定义参数对象
    url.replace(pattern, ($, $1, $2) => {
      parames[$1] = $2;
    });
    return parames;
  }
  console.log(queryURLParams(url))
</script>

<script>
  let URL = "http://www.baidu.com?name=elephant&age=25&sex=male&num=100#smallpig"
  function queryURLParams(url) {
    // 1.创建a标签
    let link = document.createElement('a');
    link.href = url;
    let searchUrl = link.search.substr(1); // 获取问号后面字符串
    let hashUrl = link.hash.substr(1); // 获取#后面的值
    let obj = {}; // 声明参数对象
    // 2.向对象中进行存储
    hashUrl ? obj['HASH'] = hashUrl : null; // #后面是否有值

    let list = searchUrl.split("&");
    for (let i = 0; i < list.length; i++) {
      let arr = list[i].split("=");
      obj[arr[0]] = arr[1];
    }
    return obj;
  }
  console.log(queryURLParams(URL))
</script>

<script>
  let URL = "http://www.baidu.com?name=elephant&age=25&sex=male&num=100"
  function queryURLParams(URL) {
    // const url = location.search; // 项目中可直接通过search方法获取url中"?"符后的字串
    let url = URL.split("?")[1];
    let obj = {}; // 声明参数对象
    let arr = url.split("&"); // 以&符号分割为数组
    for (let i = 0; i < arr.length; i++) {
      let arrNew = arr[i].split("="); // 以"="分割为数组
      obj[arrNew[0]] = arrNew[1];
    }
    return obj;
  }
  console.log(queryURLParams(URL))
</script>

<script>
  let URL = "http://www.baidu.com?name=elephant&age=25&sex=male&num=100"
  function queryURLParams(URL) {
    let url = URL.split("?")[1];
    const urlSearchParams = new URLSearchParams(url);
    const params = Object.fromEntries(urlSearchParams.entries());
    return params
  }
  console.log(queryURLParams(URL))
</script>

getUrlParamValue = function (name) {
if (name == null || name == 'undefined') {return null; }
var searchStr = decodeURI(location.search);
var infoIndex = searchStr.indexOf(name + "=");
if (infoIndex == -1) { return null; }
var searchInfo = searchStr.substring(infoIndex + name.length + 1);
var tagIndex = searchInfo.indexOf("&");
if (tagIndex！= -1) { searchInfo = searchInfo.substring(0, tagIndex); }
return searchInfo;
};