如何使用 Javascript 和正则表达式解析 url？

Question

我想解析一些具有以下格式的网址：-

var url ="http://www.example.com/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p&mycracker=ch_vn_clothing_subcategory_Puma&ref=b41c8097-8efe-4acf-8919-0fa81bcb590a" 

所有网址的域名和其他部分不必相同，它们可以有所不同，即我正在寻找一个通用的解决方案。

基本上我想去掉所有其他的东西，只得到一部分：

/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p

我想用 JavaScript 和正则表达式来解析这个

我是这样做的：

var mapObj = {"/^(http:\/\/)?.*?\//":"","(&mycracker.+)":"","(&ref.+)":""};
var re = new RegExp(Object.keys(mapObj).join("|"),"gi");
url = url.replace(re, function(matched){
  return mapObj[matched];
}); 

但它返回这个

http://www.example.com/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43pundefined

我在哪里做的不对？还是有其他更简单的解决方案？

Answer 1

var url ="http://www.example.com/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p&mycracker=ch_vn_clothing_subcategory_Puma&ref=b41c8097-8efe-4acf-8919-0fa81bcb590a";
var newAddr = url.substr(22,url.length);
// newAddr == "/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p&mycracker=ch_vn_clothing_subcategory_Puma&ref=b41c8097-8efe-4acf-8919-0fa81bcb590a"

22 是开始分割字符串的位置。

url.length 是要包含多少。

只要链接上的域名保持不变，此方法就有效。

Answer 2

为什么不只映射一个拆分数组？

您不需要对 URL 进行正则表达式，但您必须在循环内运行 if 语句以从中删除特定的 GET 参数。在这种特殊情况下（关键字特别）你只需要子串直到indexOf "&mycracker"

var url ="http://www.example.com/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p&mycracker=ch_vn_clothing_subcategory_Puma&ref=b41c8097-8efe-4acf-8919-0fa81bcb590a" 
var x = url.split("/");
var y = [];
x.map(function(data,index) { if (index >= 3) y.push(data); });
var path = "/"+y.join("/");
path = path.substring(0,path.indexOf("&mycracker"));

Answer 3

你可以使用：

/(?:https?:\/\/[^\/]*)(\/.*?)(?=\&mycracker)/

代码：

var s="http://www.example.com/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p&mycracker=ch_vn_clothing_subcategory_Puma&ref=b41c8097-8efe-4acf-8919-0fa81bcb590a";
var ss=/(?:https?:\/\/[^\/]*)(\/.*?)(?=\&mycracker)/;
console.log(s.match(ss)[1]);

Demo

Fiddle Demo

说明：

Answer 4

下面的代码稍微改一下，就可以取回任意参数了：

var url = "http://www.example.com/cooks/cooking-dress-wine/~no-order/pr?p%5B%5D=sort%3Dfeatured&sid=bks%2C43p&mycracker=ch_vn_clothing_subcategory_Puma&ref=b41c8097-8efe-4acf-8919-0fa81bcb590a"
var re = new RegExp(/http:\/\/[^?]+/);
var part1 = url.match(re);
var remain = url.replace(re, '');
//alert('Part1: ' + part1);
var rf = remain.split('&');
// alert('Part2: ' + rf);
var part2 = '';
for (var i = 0; i < rf.length; i++) 
    if (rf[i].match(/(p%5B%5D|sid)=/))
        part2 += rf[i] + '&';
part2 = part2.replace(/&$/, '');
//alert(part2)
url = part1 + part2;
alert(url);