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
它处理这样的URL:
空 没有任何参数 已经有一些参数 有什么?在最后，但同时没有任何参数
它不处理这样的URL:
带有片段标识符(即散列，#) 如果URL已经有必要的查询参数(那么将会有重复)
适用于:
Chrome 32 + 火狐26 + 野生动物园7.1 +
function appendQueryParameter(url, name, value) {
if (url.length === 0) {
return;
}
let rawURL = url;
// URL with `?` at the end and without query parameters
// leads to incorrect result.
if (rawURL.charAt(rawURL.length - 1) === "?") {
rawURL = rawURL.slice(0, rawURL.length - 1);
}
const parsedURL = new URL(rawURL);
let parameters = parsedURL.search;
parameters += (parameters.length === 0) ? "?" : "&";
parameters = (parameters + name + "=" + value);
return (parsedURL.origin + parsedURL.pathname + parameters);
}
使用ES6模板字符串的版本。
适用于:
Chrome 41 + 火狐32 + 野生动物园9.1 +
function appendQueryParameter(url, name, value) {
if (url.length === 0) {
return;
}
let rawURL = url;
// URL with `?` at the end and without query parameters
// leads to incorrect result.
if (rawURL.charAt(rawURL.length - 1) === "?") {
rawURL = rawURL.slice(0, rawURL.length - 1);
}
const parsedURL = new URL(rawURL);
let parameters = parsedURL.search;
parameters += (parameters.length === 0) ? "?" : "&";
parameters = `${parameters}${name}=${value}`;
return `${parsedURL.origin}${parsedURL.pathname}${parameters}`;
}
其他回答
你可以使用URLSearchParams
const urlParams = new URLSearchParams(window.location.search);
urlParams.set('order', 'date');
window.location.search = urlParams;
.第一个农业是关键，第二个是价值。
注意:这在任何版本的ie浏览器中都不支持(但在Edge中支持)
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
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;
}
}
});
我添加我的解决方案，因为它支持相对url除了绝对url。在其他方面，它与顶部的答案相同，后者也使用Web API。
/**
* updates a relative or absolute
* by setting the search query with
* the passed key and value.
*/
export const setQueryParam = (url, key, value) => {
const dummyBaseUrl = 'https://dummy-base-url.com';
const result = new URL(url, dummyBaseUrl);
result.searchParams.set(key, value);
return result.toString().replace(dummyBaseUrl, '');
};
还有人开玩笑说:
// some jest tests
describe('setQueryParams', () => {
it('sets param on relative url with base path', () => {
// act
const actual = setQueryParam(
'/', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('/?ref=some-value');
});
it('sets param on relative url with no path', () => {
// act
const actual = setQueryParam(
'', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('/?ref=some-value');
});
it('sets param on relative url with some path', () => {
// act
const actual = setQueryParam(
'/some-path', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('/some-path?ref=some-value');
});
it('overwrites existing param', () => {
// act
const actual = setQueryParam(
'/?ref=prev-value', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('/?ref=some-value');
});
it('sets param while another param exists', () => {
// act
const actual = setQueryParam(
'/?other-param=other-value', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('/?other-param=other-value&ref=some-value');
});
it('honors existing base url', () => {
// act
const actual = setQueryParam(
'https://base.com', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('https://base.com/?ref=some-value');
});
it('honors existing base url with some path', () => {
// act
const actual = setQueryParam(
'https://base.com/some-path', 'ref', 'some-value',
);
// assert
expect(actual).toEqual('https://base.com/some-path?ref=some-value');
});
});
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…
