是否存在用于解析 URL 路径段(矩阵）参数的 JavaScript 库

>我最近写了一个Node.js中间件来解析矩阵参数。我已经指定了它遵循的规则以及它生成的输出的格式。

例如，您的应用程序.js如下所示：

let app = require ('express') (),
  matrixParser = require ('matrix-parser');
app.use (matrixParser ());
app.get ('/cars*', (req, res) => {
  //notice the asterisk after '/cars'
  console.log (JSON.stringify (req.matrix, null, 2));
  res.send ('Thanks=)');
});
app.listen (9090);

和你的 URI 如下所示：

http://localhost:9090/cars;color=red;make=Tesla?page=1&perPage=10

然后，您可以使用 curl 测试矩阵解析器功能，如下所示：

curl "http://localhost:9090/cars;color=red;make=Tesla?page=1&perPage=10"

然后req.matrix设置为以下对象：

[
  {
    "segment": "cars",
    "matrix": {
      "color": "red",
      "make": "Tesla"
    }
  }
]

查询字符串（页面，per_page）保持不变（您可以通过简单地编写req.query来查看）

在这一点上写答案可能为时已晚，但将来仍然可能派上用场。

下面是存储库：https://github.com/duaraghav8/matrix-parser

npm install matrix-parser

编辑：很抱歉没有早点提供更详细的代码答案，这是我对SO的第一个贡献，我会花一些时间来掌握它。

我找到了URI.js。但是，如果您不想使用该库，我认为此函数将完成您正在寻找的功能（不太确定decodeURIComponent）：

var urlString = "http://somehost:9090/cars;color=red;make=Tesla?page=1&perPage=10"
var getParams = function (urlString) {
    return decodeURIComponent(urlString) // decode the URL (?)
        .match(/'/((?!.+'/).+)'?/)
            // the regex looks for a slash that is NOT
            // followed by at least one character and eventually another slash
            // given var urlString = "http://somehost:9090/cars;color=red;make=Tesla?page=1&perPage=10"
            // we don't want        -------^              ^                         ^
            // we want this slash                   ------|                         |
            // all the way until this question mark --------------------------------|
            // regex explanation:
            /*
                '/                  first slash
                (                   open capturing group
                    (?!             lookbehind for NOT
                        .+'/        any character followed by a slash (/)
                    )
                    .+              capture one or more characters (greedy) past
                )                   the close of the capturing group and until
                '?                  a question mark
            */
        [1] // match will return two groups, which will look like:
            // ["/cars;color=red;make=Tesla?", "cars;color=red;make=Tesla"]
            // we want the second one (otherwise we'd have to .slice(1,-1) the string)
        .split(";")
            // split it at the semicolons
            // if you know you're always going to have "name" followed by a semicolon,
            // you might consider using .slice(1) on this part, so you can get rid of 
            // the if statement below (still keep the p[c[0]] = c[1] part though )
        .reduce(function (p, c) {
            // split it at the equals sign for a key/value in indices 0 and 1
            c = c.split("=");
            // if the length is greater than one, aka we have a key AND a value
            // e.g., c == ["color", "red"]
            if (c.length > 1) {
                // give the previous object a key of c[0] equal to c[1]
                // i.e., p["color"] = "red"
                p[c[0]] = c[1];
            }
            return p; // return p, so that we can keep adding keys to the object
        }, {}); // we pass an object, which will act as p on the first call
}
console.log(getParams(urlString));  // { color: "red", make: "Tesla" }

除了正则表达式，您还可以使用我在上面的评论中发布的内容：

urlString.split("?")[0].split("/").pop().split(";").reduce( /* etc */)

现在我想要一辆特斯拉...