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。
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
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);
}
其他回答
试一试 正则表达式,如此之慢,因此:
var SetParamUrl = function(_k, _v) {// replace and add new parameters
let arrParams = window.location.search !== '' ? decodeURIComponent(window.location.search.substr(1)).split('&').map(_v => _v.split('=')) : Array();
let index = arrParams.findIndex((_v) => _v[0] === _k);
index = index !== -1 ? index : arrParams.length;
_v === null ? arrParams = arrParams.filter((_v, _i) => _i != index) : arrParams[index] = [_k, _v];
let _search = encodeURIComponent(arrParams.map(_v => _v.join('=')).join('&'));
let newurl = window.location.protocol + "//" + window.location.host + window.location.pathname + (arrParams.length > 0 ? '?' + _search : '');
// window.location = newurl; //reload
if (history.pushState) { // without reload
window.history.pushState({path:newurl}, null, newurl);
}
};
var GetParamUrl = function(_k) {// get parameter by key
let sPageURL = decodeURIComponent(window.location.search.substr(1)),
sURLVariables = sPageURL.split('&').map(_v => _v.split('='));
let _result = sURLVariables.find(_v => _v[0] === _k);
return _result[1];
};
例子:
// https://some.com/some_path
GetParamUrl('cat');//undefined
SetParamUrl('cat', "strData");// https://some.com/some_path?cat=strData
GetParamUrl('cat');//strData
SetParamUrl('sotr', "strDataSort");// https://some.com/some_path?cat=strData&sotr=strDataSort
GetParamUrl('sotr');//strDataSort
SetParamUrl('cat', "strDataTwo");// https://some.com/some_path?cat=strDataTwo&sotr=strDataSort
GetParamUrl('cat');//strDataTwo
//remove param
SetParamUrl('cat', null);// https://some.com/some_path?sotr=strDataSort
你可以使用其中一个:
https://developer.mozilla.org/en-US/docs/Web/API/URL https://developer.mozilla.org/en/docs/Web/API/URLSearchParams
例子:
var url = new URL("http://foo.bar/?x=1&y=2");
// If your expected result is "http://foo.bar/?x=1&y=2&x=42"
url.searchParams.append('x', 42);
// If your expected result is "http://foo.bar/?x=42&y=2"
url.searchParams.set('x', 42);
你可以使用url。href或URL . tostring()来获取完整的URL
如果你有一个url字符串,你想用一个参数来装饰,你可以试试这个在线程序:
urlstring += ( urlstring.match( /[\?]/g ) ? '&' : '?' ) + 'param=value';
这意味着什么?将是参数的前缀,但如果已经有?在urlstring中,than &将是前缀。
我也会建议做encodeURI(paramvariable),如果你没有硬编码参数,但它是在一个paramvariable;或者里面有有趣的角色。
encodeURI函数的使用请参见javascript URL编码。
我有一个'类',这是:
function QS(){
this.qs = {};
var s = location.search.replace( /^\?|#.*$/g, '' );
if( s ) {
var qsParts = s.split('&');
var i, nv;
for (i = 0; i < qsParts.length; i++) {
nv = qsParts[i].split('=');
this.qs[nv[0]] = nv[1];
}
}
}
QS.prototype.add = function( name, value ) {
if( arguments.length == 1 && arguments[0].constructor == Object ) {
this.addMany( arguments[0] );
return;
}
this.qs[name] = value;
}
QS.prototype.addMany = function( newValues ) {
for( nv in newValues ) {
this.qs[nv] = newValues[nv];
}
}
QS.prototype.remove = function( name ) {
if( arguments.length == 1 && arguments[0].constructor == Array ) {
this.removeMany( arguments[0] );
return;
}
delete this.qs[name];
}
QS.prototype.removeMany = function( deleteNames ) {
var i;
for( i = 0; i < deleteNames.length; i++ ) {
delete this.qs[deleteNames[i]];
}
}
QS.prototype.getQueryString = function() {
var nv, q = [];
for( nv in this.qs ) {
q[q.length] = nv+'='+this.qs[nv];
}
return q.join( '&' );
}
QS.prototype.toString = QS.prototype.getQueryString;
//examples
//instantiation
var qs = new QS;
alert( qs );
//add a sinle name/value
qs.add( 'new', 'true' );
alert( qs );
//add multiple key/values
qs.add( { x: 'X', y: 'Y' } );
alert( qs );
//remove single key
qs.remove( 'new' )
alert( qs );
//remove multiple keys
qs.remove( ['x', 'bogus'] )
alert( qs );
我已经重写了toString方法,所以不需要调用QS::getQueryString,你可以使用QS::toString,或者像我在示例中所做的那样,仅仅依赖于对象被强制转换为字符串。
好的,在这里我比较两个函数,一个由我自己(regExp)和另一个由(annakata)。
将数组:
function insertParam(key, value)
{
key = escape(key); value = escape(value);
var kvp = document.location.search.substr(1).split('&');
var i=kvp.length; var x; while(i--)
{
x = kvp[i].split('=');
if (x[0]==key)
{
x[1] = value;
kvp[i] = x.join('=');
break;
}
}
if(i<0) {kvp[kvp.length] = [key,value].join('=');}
//this will reload the page, it's likely better to store this until finished
return "&"+kvp.join('&');
}
正则表达式的方法:
function addParameter(param, value)
{
var regexp = new RegExp("(\\?|\\&)" + param + "\\=([^\\&]*)(\\&|$)");
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…