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
你需要适应的基本实现是这样的:
function insertParam(key, value) {
key = encodeURIComponent(key);
value = encodeURIComponent(value);
// kvp looks like ['key1=value1', 'key2=value2', ...]
var kvp = document.location.search.substr(1).split('&');
let i=0;
for(; i<kvp.length; i++){
if (kvp[i].startsWith(key + '=')) {
let pair = kvp[i].split('=');
pair[1] = value;
kvp[i] = pair.join('=');
break;
}
}
if(i >= kvp.length){
kvp[kvp.length] = [key,value].join('=');
}
// can return this or...
let params = kvp.join('&');
// reload page with new params
document.location.search = params;
}
这大约是正则表达式或基于搜索的解决方案的两倍，但这完全取决于查询字符串的长度和任何匹配的索引
为了完成起见，我以慢速regex方法为基准(大约慢了150%)
function insertParam2(key,value)
{
key = encodeURIComponent(key); value = encodeURIComponent(value);
var s = document.location.search;
var kvp = key+"="+value;
var r = new RegExp("(&|\\?)"+key+"=[^\&]*");
s = s.replace(r,"$1"+kvp);
if(!RegExp.$1) {s += (s.length>0 ? '&' : '?') + kvp;};
//again, do what you will here
document.location.search = s;
}
其他回答
我是这么做的。使用我的editParams()函数，你可以添加，删除或更改任何参数，然后使用内置的replaceState()函数更新URL:
window.history.replaceState('object or string', 'Title', 'page.html' + editParams('enable', 'true'));
// background functions below:
// add/change/remove URL parameter
// use a value of false to remove parameter
// returns a url-style string
function editParams (key, value) {
key = encodeURI(key);
var params = getSearchParameters();
if (Object.keys(params).length === 0) {
if (value !== false)
return '?' + key + '=' + encodeURI(value);
else
return '';
}
if (value !== false)
params[key] = encodeURI(value);
else
delete params[key];
if (Object.keys(params).length === 0)
return '';
return '?' + $.map(params, function (value, key) {
return key + '=' + value;
}).join('&');
}
// Get object/associative array of URL parameters
function getSearchParameters () {
var prmstr = window.location.search.substr(1);
return prmstr !== null && prmstr !== "" ? transformToAssocArray(prmstr) : {};
}
// convert parameters from url-style string to associative array
function transformToAssocArray (prmstr) {
var params = {},
prmarr = prmstr.split("&");
for (var i = 0; i < prmarr.length; i++) {
var tmparr = prmarr[i].split("=");
params[tmparr[0]] = tmparr[1];
}
return params;
}
有时我们看到?在URL结尾，我找到了一些解决方案，生成的结果为file.php?&foo=bar。我想出了我自己的解决方案，以完美地工作!
location.origin + location.pathname + location.search + (location.search=='' ? '?' : '&') + 'lang=ar'
注意:位置。origin不能在IE中工作，这里是它的修复。
好的，在这里我比较两个函数，一个由我自己(regExp)和另一个由(annakata)。
将数组:
function insertParam(key, value)
{
key = escape(key); value = escape(value);
var kvp = document.location.search.substr(1).split('&');
var i=kvp.length; var x; while(i--)
{
x = kvp[i].split('=');
if (x[0]==key)
{
x[1] = value;
kvp[i] = x.join('=');
break;
}
}
if(i<0) {kvp[kvp.length] = [key,value].join('=');}
//this will reload the page, it's likely better to store this until finished
return "&"+kvp.join('&');
}
正则表达式的方法:
function addParameter(param, value)
{
var regexp = new RegExp("(\\?|\\&)" + param + "\\=([^\\&]*)(\\&|$)");
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…
你需要适应的基本实现是这样的:
function insertParam(key, value) {
key = encodeURIComponent(key);
value = encodeURIComponent(value);
// kvp looks like ['key1=value1', 'key2=value2', ...]
var kvp = document.location.search.substr(1).split('&');
let i=0;
for(; i<kvp.length; i++){
if (kvp[i].startsWith(key + '=')) {
let pair = kvp[i].split('=');
pair[1] = value;
kvp[i] = pair.join('=');
break;
}
}
if(i >= kvp.length){
kvp[kvp.length] = [key,value].join('=');
}
// can return this or...
let params = kvp.join('&');
// reload page with new params
document.location.search = params;
}
这大约是正则表达式或基于搜索的解决方案的两倍，但这完全取决于查询字符串的长度和任何匹配的索引
为了完成起见，我以慢速regex方法为基准(大约慢了150%)
function insertParam2(key,value)
{
key = encodeURIComponent(key); value = encodeURIComponent(value);
var s = document.location.search;
var kvp = key+"="+value;
var r = new RegExp("(&|\\?)"+key+"=[^\&]*");
s = s.replace(r,"$1"+kvp);
if(!RegExp.$1) {s += (s.length>0 ? '&' : '?') + kvp;};
//again, do what you will here
document.location.search = s;
}
以下功能将帮助您添加，更新和删除参数或从URL。
/ / example1and
var myURL = '/search';
myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?location=california
myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?location=new%20york
myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search
/ / example2
var myURL = '/search?category=mobile';
myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?category=mobile&location=california
myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?category=mobile&location=new%20york
myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search?category=mobile
/ /青年们
var myURL = '/search?location=texas';
myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?location=california
myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?location=new%20york
myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search
/ / example4
var myURL = '/search?category=mobile&location=texas';
myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?category=mobile&location=california
myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?category=mobile&location=new%20york
myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search?category=mobile
/ / example5
var myURL = 'https://example.com/search?location=texas#fragment';
myURL = updateUrl(myURL,'location','california');
console.log('added location...' + myURL);
//added location.../search?location=california#fragment
myURL = updateUrl(myURL,'location','new york');
console.log('updated location...' + myURL);
//updated location.../search?location=new%20york#fragment
myURL = updateUrl(myURL,'location');
console.log('removed location...' + myURL);
//removed location.../search#fragment
这是函数。
function updateUrl(url,key,value){
if(value!==undefined){
value = encodeURI(value);
}
var hashIndex = url.indexOf("#")|0;
if (hashIndex === -1) hashIndex = url.length|0;
var urls = url.substring(0, hashIndex).split('?');
var baseUrl = urls[0];
var parameters = '';
var outPara = {};
if(urls.length>1){
parameters = urls[1];
}
if(parameters!==''){
parameters = parameters.split('&');
for(k in parameters){
var keyVal = parameters[k];
keyVal = keyVal.split('=');
var ekey = keyVal[0];
var evalue = '';
if(keyVal.length>1){
evalue = keyVal[1];
}
outPara[ekey] = evalue;
}
}
if(value!==undefined){
outPara[key] = value;
}else{
delete outPara[key];
}
parameters = [];
for(var k in outPara){
parameters.push(k + '=' + outPara[k]);
}
var finalUrl = baseUrl;
if(parameters.length>0){
finalUrl += '?' + parameters.join('&');
}
return finalUrl + url.substring(hashIndex);
}
