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
这是我自己的尝试，但我将使用annakata的答案，因为它看起来更清晰:
function AddUrlParameter(sourceUrl, parameterName, parameterValue, replaceDuplicates)
{
if ((sourceUrl == null) || (sourceUrl.length == 0)) sourceUrl = document.location.href;
var urlParts = sourceUrl.split("?");
var newQueryString = "";
if (urlParts.length > 1)
{
var parameters = urlParts[1].split("&");
for (var i=0; (i < parameters.length); i++)
{
var parameterParts = parameters[i].split("=");
if (!(replaceDuplicates && parameterParts[0] == parameterName))
{
if (newQueryString == "")
newQueryString = "?";
else
newQueryString += "&";
newQueryString += parameterParts[0] + "=" + parameterParts[1];
}
}
}
if (newQueryString == "")
newQueryString = "?";
else
newQueryString += "&";
newQueryString += parameterName + "=" + parameterValue;
return urlParts[0] + newQueryString;
}
另外，我从stackoverflow上的另一篇文章中找到了这个jQuery插件，如果你需要更多的灵活性，你可以使用它: http://plugins.jquery.com/project/query-object
我认为代码应该是(还没有测试):
return $.query.parse(sourceUrl).set(parameterName, parameterValue).toString();
其他回答
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);
}
我喜欢穆罕穆德·法提赫·耶尔达兹的回答，即使他没有回答整个问题。
在他回答的同一行中，我使用了这样的代码:
“它不控制参数的存在，也不改变现有的值。它把你的参数加到最后"
/** add a parameter at the end of the URL. Manage '?'/'&', but not the existing parameters.
* does escape the value (but not the key)
*/
function addParameterToURL(_url,_key,_value){
var param = _key+'='+escape(_value);
var sep = '&';
if (_url.indexOf('?') < 0) {
sep = '?';
} else {
var lastChar=_url.slice(-1);
if (lastChar == '&') sep='';
if (lastChar == '?') sep='';
}
_url += sep + param;
return _url;
}
测试者:
/*
function addParameterToURL_TESTER_sub(_url,key,value){
//log(_url);
log(addParameterToURL(_url,key,value));
}
function addParameterToURL_TESTER(){
log('-------------------');
var _url ='www.google.com';
addParameterToURL_TESTER_sub(_url,'key','value');
addParameterToURL_TESTER_sub(_url,'key','Text Value');
_url ='www.google.com?';
addParameterToURL_TESTER_sub(_url,'key','value');
_url ='www.google.com?A=B';
addParameterToURL_TESTER_sub(_url,'key','value');
_url ='www.google.com?A=B&';
addParameterToURL_TESTER_sub(_url,'key','value');
_url ='www.google.com?A=1&B=2';
addParameterToURL_TESTER_sub(_url,'key','value');
}//*/
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
这将在所有现代浏览器中工作。
function insertParam(key,value) {
if (history.pushState) {
var newurl = window.location.protocol + "//" + window.location.host + window.location.pathname + '?' +key+'='+value;
window.history.pushState({path:newurl},'',newurl);
}
}
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;
}
