是否有一种通过jQuery(或不使用)检索查询字符串值的无插件方法?
如果是,怎么办?如果没有,是否有插件可以这样做?
是否有一种通过jQuery(或不使用)检索查询字符串值的无插件方法?
如果是,怎么办?如果没有,是否有插件可以这样做?
当前回答
Node.js的源代码中有一个健壮的实现https://github.com/joyent/node/blob/master/lib/querystring.js
TJ的qs也执行嵌套参数解析https://github.com/visionmedia/node-querystring
其他回答
我认为这是实现这一点的准确和简洁的方法(修改自http://css-tricks.com/snippets/javascript/get-url-variables/):
function getQueryVariable(variable) {
var query = window.location.search.substring(1), // Remove the ? from the query string.
vars = query.split("&"); // Split all values by ampersand.
for (var i = 0; i < vars.length; i++) { // Loop through them...
var pair = vars[i].split("="); // Split the name from the value.
if (pair[0] == variable) { // Once the requested value is found...
return ( pair[1] == undefined ) ? null : pair[1]; // Return null if there is no value (no equals sign), otherwise return the value.
}
}
return undefined; // Wasn't found.
}
Node.js的源代码中有一个健壮的实现https://github.com/joyent/node/blob/master/lib/querystring.js
TJ的qs也执行嵌套参数解析https://github.com/visionmedia/node-querystring
var getUrlParameters = function (name, url) {
if (!name) {
return undefined;
}
name = name.replace(/[\[]/, '\\[').replace(/[\]]/, '\\]');
url = url || location.search;
var regex = new RegExp('[\\?&#]' + name + '=?([^&#]*)', 'gi'), result, resultList = [];
while (result = regex.exec(url)) {
resultList.push(decodeURIComponent(result[1].replace(/\+/g, ' ')));
}
return resultList.length ? resultList.length === 1 ? resultList[0] : resultList : undefined;
};
快速、轻松、快速:
功能:
function getUrlVar() {
var result = {};
var location = window.location.href.split('#');
var parts = location[0].replace(/[?&]+([^=&]+)=([^&]*)/gi, function(m,key,value) {
result [key] = value;
});
return result;
}
用法:
var varRequest = getUrlVar()["theUrlVarName"];
我喜欢这个(摘自jqueryhowto.blogspot.co.uk):
// get an array with all querystring values
// example: var valor = getUrlVars()["valor"];
function getUrlVars() {
var vars = [], hash;
var hashes = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&');
for (var i = 0; i < hashes.length; i++) {
hash = hashes[i].split('=');
vars.push(hash[0]);
vars[hash[0]] = hash[1];
}
return vars;
}
对我来说很棒。