如何从JavaScript检索GET参数

考虑:

http://example.com/page.html?returnurl=%2Fadmin

对于page.html内的js，它如何检索GET参数?

对于上面的简单例子，func('returnurl')应该是/admin。

但它也应该适用于复杂的查询字符串…

当前回答

我这样做(检索一个特定的get-parameter，这里'parameterName'):

var parameterValue = decodeURIComponent(window.location.search.match(/(\?|&)parameterName\=([^&]*)/)[2]);

2014-11-18 13:02:06

其他回答

var getQueryParam = function(param) {
    var found;
    window.location.search.substr(1).split("&").forEach(function(item) {
        if (param ==  item.split("=")[0]) {
            found = item.split("=")[1];
        }
    });
    return found;
};

2015-04-14 12:25:28

我创建了一个简单的JavaScript函数来从URL访问GET参数。

只要包含这个JavaScript源代码，就可以访问get参数。例如:在http://example.com/index.php?language=french中，语言变量可以通过$_GET["language"]访问。类似地，所有参数的列表将作为数组存储在变量$_GET_Params中。JavaScript和HTML都在下面的代码片段中提供:

<!DOCTYPE html> <html> <body>  <script> function $_GET() { // Get the Full href of the page e.g. http://www.google.com/files/script.php?v=1.8.7&country=india var href = window.location.href; // Get the protocol e.g. http var protocol = window.location.protocol + "//"; // Get the host name e.g. www.google.com var hostname = window.location.hostname; // Get the pathname e.g. /files/script.php var pathname = window.location.pathname; // Remove protocol part var queries = href.replace(protocol, ''); // Remove host part queries = queries.replace(hostname, ''); // Remove pathname part queries = queries.replace(pathname, ''); // Presently, what is left in the variable queries is : ?v=1.8.7&country=india // Perform query functions if present if (queries != "" && queries != "?") { // Remove question mark '?' queries = queries.slice(1); // Split all the different queries queries = queries.split("&"); // Get the number of queries var length = queries.length; // Declare global variables to store keys and elements $_GET_Params = new Array(); $_GET = {}; // Perform functions per query for (var i = 0; i < length; i++) { // Get the present query var key = queries[i]; // Split the query and the value key = key.split("="); // Assign value to the $_GET variable $_GET[key[0]] = [key[1]]; // Assign value to the $_GET_Params variable $_GET_Params[i] = key[0]; } } } // Execute the function $_GET(); </script> <h1>GET Parameters</h1> <h2>Try to insert some get parameter and access it through JavaScript</h2> </body> </html>

2016-06-12 02:58:38

这是另一个基于Kat和Bakudan例子的例子，但让它更通用一点。

function getParams ()
{
    var result = {};
    var tmp = [];

    location.search
        .substr (1)
        .split ("&")
        .forEach (function (item)
        {
            tmp = item.split ("=");
            result [tmp[0]] = decodeURIComponent (tmp[1]);
        });

    return result;
}

location.getParams = getParams;

console.log (location.getParams());
console.log (location.getParams()["returnurl"]);

2016-11-30 22:02:31

你应该使用URL和URLSearchParams本地函数:

let url = new url ("https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8&q=mdn%20query%20string") let params = new URLSearchParams(url.search); Let sourceid = params.get('sourceid') // 'chrome-instant' Let q = params.get('q') // 'mdn查询字符串' 让ie = params.has('ie') // true params.append(“平”、“乒乓球”) console.log (sourceid) console.log (q) console.log (ie) console.log (params.toString ()) console.log (params.get(“平”))

https://developer.mozilla.org/en-US/docs/Web/API/URLSearchParams https://polyfill.io/v2/docs/features/

2016-09-21 15:08:37

用窗户。位置的对象。这段代码提供了不带问号的GET。

window.location.search.substr(1)

在您的示例中，它将返回returnurl=%2Fadmin

编辑:我擅自改变了Qwerty的答案，这真的很好，正如他指出的那样，我完全按照OP的要求做了:

function findGetParameter(parameterName) {
    var result = null,
        tmp = [];
    location.search
        .substr(1)
        .split("&")
        .forEach(function (item) {
          tmp = item.split("=");
          if (tmp[0] === parameterName) result = decodeURIComponent(tmp[1]);
        });
    return result;
}

我从他的代码中删除了重复的函数执行，替换为一个变量(tmp)，还添加了decodeURIComponent，完全符合OP的要求。我不确定这是不是安全问题。

或者使用普通的for循环，即使在IE8中也能工作:

function findGetParameter(parameterName) {
    var result = null,
        tmp = [];
    var items = location.search.substr(1).split("&");
    for (var index = 0; index < items.length; index++) {
        tmp = items[index].split("=");
        if (tmp[0] === parameterName) result = decodeURIComponent(tmp[1]);
    }
    return result;
}

2011-03-27 10:17:32

如何从JavaScript检索GET参数

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