在一个使用AJAX调用的web应用程序中,我需要提交一个请求,但在URL的末尾添加一个参数,例如:

原始URL:

http://server/myapp.php?id=10

导致的网址:

http://server/myapp.php?id=10&enabled=true

寻找一个JavaScript函数,该函数解析URL并查看每个参数,然后添加新参数或更新已经存在的值。


当前回答

加入@Vianney的回答https://stackoverflow.com/a/44160941/6609678

我们可以导入node中的内置URL模块,如下所示

const { URL } = require('url');

例子:

Terminal $ node
> const { URL } = require('url');
undefined
> let url = new URL('', 'http://localhost:1989/v3/orders');
undefined
> url.href
'http://localhost:1989/v3/orders'
> let fetchAll=true, timePeriod = 30, b2b=false;
undefined
> url.href
'http://localhost:1989/v3/orders'
>  url.searchParams.append('fetchAll', fetchAll);
undefined
>  url.searchParams.append('timePeriod', timePeriod);
undefined
>  url.searchParams.append('b2b', b2b);
undefined
> url.href
'http://localhost:1989/v3/orders?fetchAll=true&timePeriod=30&b2b=false'
> url.toString()
'http://localhost:1989/v3/orders?fetchAll=true&timePeriod=30&b2b=false'

有用的链接:

https://developer.mozilla.org/en-US/docs/Web/API/URL https://developer.mozilla.org/en/docs/Web/API/URLSearchParams

其他回答

好的,在这里我比较两个函数,一个由我自己(regExp)和另一个由(annakata)。

将数组:

function insertParam(key, value)
{
    key = escape(key); value = escape(value);

    var kvp = document.location.search.substr(1).split('&');

    var i=kvp.length; var x; while(i--) 
    {
        x = kvp[i].split('=');

        if (x[0]==key)
        {
                x[1] = value;
                kvp[i] = x.join('=');
                break;
        }
    }

    if(i<0) {kvp[kvp.length] = [key,value].join('=');}

    //this will reload the page, it's likely better to store this until finished
    return "&"+kvp.join('&'); 
}

正则表达式的方法:

function addParameter(param, value)
{
    var regexp = new RegExp("(\\?|\\&)" + param + "\\=([^\\&]*)(\\&|$)");
    if (regexp.test(document.location.search)) 
        return (document.location.search.toString().replace(regexp, function(a, b, c, d)
        {
                return (b + param + "=" + value + d);
        }));
    else 
        return document.location.search+ param + "=" + value;
}

测试用例:

time1=(new Date).getTime();
for (var i=0;i<10000;i++)
{
addParameter("test","test");
}
time2=(new Date).getTime();
for (var i=0;i<10000;i++)
{
insertParam("test","test");
}

time3=(new Date).getTime();

console.log((time2-time1)+" "+(time3-time2));

似乎即使使用最简单的解决方案(当regexp只使用test而不输入.replace函数时),它仍然比分裂要慢…好。Regexp有点慢,但是…喔…

var MyApp = new Class();

MyApp.extend({
    utility: {
        queryStringHelper: function (url) {
            var originalUrl = url;
            var newUrl = url;
            var finalUrl;
            var insertParam = function (key, value) {
                key = escape(key);
                value = escape(value);

                //The previous post had the substr strat from 1 in stead of 0!!!
                var kvp = newUrl.substr(0).split('&');

                var i = kvp.length;
                var x;
                while (i--) {
                    x = kvp[i].split('=');

                    if (x[0] == key) {
                        x[1] = value;
                        kvp[i] = x.join('=');
                        break;
                    }
                }

                if (i < 0) {
                    kvp[kvp.length] = [key, value].join('=');
                }

                finalUrl = kvp.join('&');

                return finalUrl;
            };

            this.insertParameterToQueryString = insertParam;

            this.insertParams = function (keyValues) {
                for (var keyValue in keyValues[0]) {
                    var key = keyValue;
                    var value = keyValues[0][keyValue];
                    newUrl = insertParam(key, value);
                }
                return newUrl;
            };

            return this;
        }
    }
});

以下几点:

合并重复的查询字符串参数 使用绝对和相对url 在浏览器和节点中工作

/**
 * Adds query params to existing URLs (inc merging duplicates)
 * @param {string} url - src URL to modify
 * @param {object} params - key/value object of params to add
 * @returns {string} modified URL
 */
function addQueryParamsToUrl(url, params) {

    // if URL is relative, we'll need to add a fake base
    var fakeBase = !url.startsWith('http') ? 'http://fake-base.com' : undefined;
    var modifiedUrl = new URL(url || '', fakeBase);

    // add/update params
    Object.keys(params).forEach(function(key) {
        if (modifiedUrl.searchParams.has(key)) {
            modifiedUrl.searchParams.set(key, params[key]);
        }
        else {
            modifiedUrl.searchParams.append(key, params[key]);
        }
    });

    // return as string (remove fake base if present)
    return modifiedUrl.toString().replace(fakeBase, '');
}

例子:

// returns /guides?tag=api
addQueryParamsToUrl('/guides?tag=hardware', { tag:'api' })

// returns https://orcascan.com/guides?tag=api
addQueryParamsToUrl('https://orcascan.com/guides?tag=hardware', { tag: 'api' })

就我所知,上面的答案都没有解决查询字符串包含参数的情况,这些参数本身就是一个数组,因此会出现不止一次,例如:

http://example.com?sizes[]=a&sizes[]=b

下面的函数是我用来更新document.location.search的。它接受一个键/值对数组作为参数,它将返回一个修改后的版本,你可以做任何你想做的事情。我是这样使用的:

var newParams = [
    ['test','123'],
    ['best','456'],
    ['sizes[]','XXL']
];
var newUrl = document.location.pathname + insertParams(newParams);
history.replaceState('', '', newUrl);

如果当前url为:

http://example.com/index.php?test=replaceme&sizes [] = XL

这会让你

http://example.com/index.php?test=123&sizes[]=XL&尺寸[]=XXL&best=456

函数

function insertParams(params) {
    var result;
    var ii = params.length;
    var queryString = document.location.search.substr(1);
    var kvps = queryString ? queryString.split('&') : [];
    var kvp;
    var skipParams = [];
    var i = kvps.length;
    while (i--) {
        kvp = kvps[i].split('=');
        if (kvp[0].slice(-2) != '[]') {
            var ii = params.length;
            while (ii--) {
                if (params[ii][0] == kvp[0]) {
                    kvp[1] = params[ii][1];
                    kvps[i] = kvp.join('=');
                    skipParams.push(ii);
                }
            }
        }
    }
    var ii = params.length;
    while (ii--) {
        if (skipParams.indexOf(ii) === -1) {
            kvps.push(params[ii].join('='));
        }
    }
    result = kvps.length ? '?' + kvps.join('&') : '';
    return result;
}

Vianney Bajart的答案是正确的;然而,URL只会工作,如果你有完整的URL端口,主机,路径和查询:

new URL('http://server/myapp.php?id=10&enabled=true')

URLSearchParams只会在你只传递查询字符串时起作用:

new URLSearchParams('?id=10&enabled=true')

如果你有一个不完整或相对的URL,不关心的基础URL,你可以通过?获取查询字符串,然后像这样连接:

function setUrlParams(url, key, value) {
  url = url.split('?');
  usp = new URLSearchParams(url[1]);
  usp.set(key, value);
  url[1] = usp.toString();
  return url.join('?');
}

let url = 'myapp.php?id=10';
url = setUrlParams(url, 'enabled', true);  // url = 'myapp.php?id=10&enabled=true'
url = setUrlParams(url, 'id', 11);         // url = 'myapp.php?id=11&enabled=true'

Internet Explorer浏览器不兼容。