我如何在javascript中解析URL到主机名和路径?

我想拿一根绳子

var a = "http://example.com/aa/bb/"

然后把它加工成一个物体

a.hostname == "example.com"

and

a.pathname == "/aa/bb"

当前回答

AngularJS的方法在这里:http://jsfiddle.net/PT5BG/4/

<!DOCTYPE html>
<html>
<head>
    <title>Parse URL using AngularJS</title>
</head>
<body ng-app ng-controller="AppCtrl" ng-init="init()">

<h3>Parse URL using AngularJS</h3>

url: <input type="text" ng-model="url" value="" style="width:780px;">

<ul>
    <li>href = {{parser.href}}</li>
    <li>protocol = {{parser.protocol}}</li>
    <li>host = {{parser.host}}</li>
    <li>hostname = {{parser.hostname}}</li>
    <li>port = {{parser.port}}</li>
    <li>pathname = {{parser.pathname}}</li>
    <li>hash = {{parser.hash}}</li>
    <li>search = {{parser.search}}</li>
</ul>

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.0.6/angular.min.js"></script>

<script>
function AppCtrl($scope) {

    $scope.$watch('url', function() {
        $scope.parser.href = $scope.url;
    });

    $scope.init = function() {
        $scope.parser = document.createElement('a');
        $scope.url = window.location;
    }

}
</script>

</body>
</html>

2013-04-15 07:02:29

其他回答

js-uri(在谷歌Code上可用)接受字符串URL并从中解析URI对象:

var some_uri = new URI("http://www.example.com/foo/bar");

alert(some_uri.authority); // www.example.com
alert(some_uri);           // http://www.example.com/foo/bar

var blah      = new URI("blah");
var blah_full = blah.resolve(some_uri);
alert(blah_full);         // http://www.example.com/foo/blah

2009-04-10 02:21:16

AngularJS的方法在这里:http://jsfiddle.net/PT5BG/4/

<!DOCTYPE html>
<html>
<head>
    <title>Parse URL using AngularJS</title>
</head>
<body ng-app ng-controller="AppCtrl" ng-init="init()">

<h3>Parse URL using AngularJS</h3>

url: <input type="text" ng-model="url" value="" style="width:780px;">

<ul>
    <li>href = {{parser.href}}</li>
    <li>protocol = {{parser.protocol}}</li>
    <li>host = {{parser.host}}</li>
    <li>hostname = {{parser.hostname}}</li>
    <li>port = {{parser.port}}</li>
    <li>pathname = {{parser.pathname}}</li>
    <li>hash = {{parser.hash}}</li>
    <li>search = {{parser.search}}</li>
</ul>

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.0.6/angular.min.js"></script>

<script>
function AppCtrl($scope) {

    $scope.$watch('url', function() {
        $scope.parser.href = $scope.url;
    });

    $scope.init = function() {
        $scope.parser = document.createElement('a');
        $scope.url = window.location;
    }

}
</script>

</body>
</html>

2013-04-15 07:02:29

使用模块模式的简单而健壮的解决方案。这包括修复IE的路径名不总是有前导正斜杠(/)。

我已经创建了一个Gist和一个JSFiddle，它提供了一个更动态的解析器。我建议你检查一下并提供反馈。

var URLParser = (function (document) {
    var PROPS = 'protocol hostname host pathname port search hash href'.split(' ');
    var self = function (url) {
        this.aEl = document.createElement('a');
        this.parse(url);
    };
    self.prototype.parse = function (url) {
        this.aEl.href = url;
        if (this.aEl.host == "") {
           this.aEl.href = this.aEl.href;
        }
        PROPS.forEach(function (prop) {
            switch (prop) {
                case 'hash':
                    this[prop] = this.aEl[prop].substr(1);
                    break;
                default:
                    this[prop] = this.aEl[prop];
            }
        }, this);
        if (this.pathname.indexOf('/') !== 0) {
            this.pathname = '/' + this.pathname;
        }
        this.requestUri = this.pathname + this.search;
    };
    self.prototype.toObj = function () {
        var obj = {};
        PROPS.forEach(function (prop) {
            obj[prop] = this[prop];
        }, this);
        obj.requestUri = this.requestUri;
        return obj;
    };
    self.prototype.toString = function () {
        return this.href;
    };
    return self;
})(document);

Demo

输出

{
 "protocol": "https:",
 "hostname": "www.example.org",
 "host": "www.example.org:5887",
 "pathname": "/foo/bar",
 "port": "5887",
 "search": "?a=1&b=2",
 "hash": "section-1",
 "href": "https://www.example.org:5887/foo/bar?a=1&b=2#section-1",
 "requestUri": "/foo/bar?a=1&b=2"
}
{
 "protocol": "ftp:",
 "hostname": "www.files.com",
 "host": "www.files.com:22",
 "pathname": "/folder",
 "port": "22",
 "search": "?id=7",
 "hash": "",
 "href": "ftp://www.files.com:22/folder?id=7",
 "requestUri": "/folder?id=7"
}

2015-01-15 13:37:22

当然，在>2016年，正确的答案是使用URL API 对于页面URL window.location And for <a href="…""> html lanchorelement API

也支持旧浏览器，使用polyfill:

<script crossorigin="anonymous" src="https://polyfill.io/v3/polyfill.min.js?features=URL"></script>

但作为另一种选择，它可以通过RegEx模式简单而准确地处理:

使用正则表达式

任何形式的

绝对/相对路径 IPv4 / IPv6 网络协议/本地文件查询/散列

返回所有URL选项，但不返回searchParams

(+)也将返回文件信息，如PHP pathInfo

2022-02-09 19:07:27

对于那些正在寻找在IE, Firefox和Chrome中工作的现代解决方案的人:

这些使用超链接元素的解决方案都不会在chrome中起同样的作用。如果你传递一个无效(或空白)的url给chrome，它总是会返回脚本被调用的主机。所以在IE中你会得到空白，而在Chrome中你会得到localhost(或其他什么)。

如果你试图看着推荐人，这是骗人的。你需要确保你返回的主机在原始url中，以处理这个问题:

    function getHostNameFromUrl(url) {
        // <summary>Parses the domain/host from a given url.</summary>
        var a = document.createElement("a");
        a.href = url;

        // Handle chrome which will default to domain where script is called from if invalid
        return url.indexOf(a.hostname) != -1 ? a.hostname : '';
    }

2013-11-06 00:20:46

我如何在javascript中解析URL到主机名和路径?

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