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
Use:
$(document).ready(function () {
var urlParams = {};
(function () {
var match,
pl = /\+/g, // Regex for replacing addition symbol with a space
search = /([^&=]+)=?([^&]*)/g,
decode = function (s) {
return decodeURIComponent(s.replace(pl, " "));
},
query = window.location.search.substring(1);
while (match = search.exec(query))
urlParams[decode(match[1])] = decode(match[2]);
})();
if (urlParams["q1"] === 1) {
return 1;
}
请检查并让我知道您的意见。另请参阅How to get querystring value using jQuery。
其他回答
最漂亮但最基本的:
data = {};
$.each(
location.search.substr(1).split('&').filter(Boolean).map(function(kvpairs){
return kvpairs.split('=')
}),
function(i,values) {
data[values.shift()] = values.join('=')
}
);
它不处理值列表,例如?a[]=1&a[]2
使用纯JavaScript和正则表达式的简单解决方案:
alert(getQueryString("p2"));
function getQueryString (Param) {
return decodeURI("http://www.example.com/?p1=p11&p2=p2222".replace(new RegExp("^(?:.*[&?]" + encodeURI(Param).replace(/[.+*]/g, "$&") + "(?:=([^&]*))?)?.*$", "i"), "$1"));
}
Js投标
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.
}
这是Andy E链接的“句柄数组样式查询字符串”版本的扩展版本。修复了一个错误(?key=1&key[]=2&key[]=3;1丢失并替换为[2,3]),进行了一些小的性能改进(重新解码值,重新计算“[”位置等),并添加了一些改进(功能化,支持?key=1&key=2,支持;分隔符)。我将变量留得很短,但添加了大量注释以使其可读(哦,我在本地函数中重用了v,如果这令人困惑,很抱歉;)。
它将处理以下查询字符串。。。
?test=Hello&pers=neek&pers[]=jeff&pers[][]=jim&pers[extra]=john&test3&nocache=13989148914891264
…把它做成一个看起来像。。。
{
"test": "Hello",
"person": {
"0": "neek",
"1": "jeff",
"2": "jim",
"length": 3,
"extra": "john"
},
"test3": "",
"nocache": "1398914891264"
}
如上所述,此版本处理一些“格式错误”数组,即-person=neek&person[]=jeff&person[]=jim或person=neek/person=jeff/person=jim,因为密钥是可识别的和有效的(至少在dotNet的NameValueCollection.Add中):
如果目标NameValueCollection中已存在指定的键例如,指定的值将添加到现有的逗号分隔的格式为“value1,value2,value3”的值列表。
似乎陪审团对重复的键有点不满意,因为没有规范。在这种情况下,多个键被存储为一个(假)数组。但请注意,我不会将基于逗号的值处理为数组。
代码:
getQueryStringKey = function(key) {
return getQueryStringAsObject()[key];
};
getQueryStringAsObject = function() {
var b, cv, e, k, ma, sk, v, r = {},
d = function (v) { return decodeURIComponent(v).replace(/\+/g, " "); }, //# d(ecode) the v(alue)
q = window.location.search.substring(1), //# suggested: q = decodeURIComponent(window.location.search.substring(1)),
s = /([^&;=]+)=?([^&;]*)/g //# original regex that does not allow for ; as a delimiter: /([^&=]+)=?([^&]*)/g
;
//# ma(make array) out of the v(alue)
ma = function(v) {
//# If the passed v(alue) hasn't been setup as an object
if (typeof v != "object") {
//# Grab the cv(current value) then setup the v(alue) as an object
cv = v;
v = {};
v.length = 0;
//# If there was a cv(current value), .push it into the new v(alue)'s array
//# NOTE: This may or may not be 100% logical to do... but it's better than loosing the original value
if (cv) { Array.prototype.push.call(v, cv); }
}
return v;
};
//# While we still have key-value e(ntries) from the q(uerystring) via the s(earch regex)...
while (e = s.exec(q)) { //# while((e = s.exec(q)) !== null) {
//# Collect the open b(racket) location (if any) then set the d(ecoded) v(alue) from the above split key-value e(ntry)
b = e[1].indexOf("[");
v = d(e[2]);
//# As long as this is NOT a hash[]-style key-value e(ntry)
if (b < 0) { //# b == "-1"
//# d(ecode) the simple k(ey)
k = d(e[1]);
//# If the k(ey) already exists
if (r[k]) {
//# ma(make array) out of the k(ey) then .push the v(alue) into the k(ey)'s array in the r(eturn value)
r[k] = ma(r[k]);
Array.prototype.push.call(r[k], v);
}
//# Else this is a new k(ey), so just add the k(ey)/v(alue) into the r(eturn value)
else {
r[k] = v;
}
}
//# Else we've got ourselves a hash[]-style key-value e(ntry)
else {
//# Collect the d(ecoded) k(ey) and the d(ecoded) sk(sub-key) based on the b(racket) locations
k = d(e[1].slice(0, b));
sk = d(e[1].slice(b + 1, e[1].indexOf("]", b)));
//# ma(make array) out of the k(ey)
r[k] = ma(r[k]);
//# If we have a sk(sub-key), plug the v(alue) into it
if (sk) { r[k][sk] = v; }
//# Else .push the v(alue) into the k(ey)'s array
else { Array.prototype.push.call(r[k], v); }
}
}
//# Return the r(eturn value)
return r;
};
这将从URL字符串中解析变量AND数组。它既不使用正则表达式,也不使用任何外部库。
function url2json(url) {
var obj={};
function arr_vals(arr){
if (arr.indexOf(',') > 1){
var vals = arr.slice(1, -1).split(',');
var arr = [];
for (var i = 0; i < vals.length; i++)
arr[i]=vals[i];
return arr;
}
else
return arr.slice(1, -1);
}
function eval_var(avar){
if (!avar[1])
obj[avar[0]] = '';
else
if (avar[1].indexOf('[') == 0)
obj[avar[0]] = arr_vals(avar[1]);
else
obj[avar[0]] = avar[1];
}
if (url.indexOf('?') > -1){
var params = url.split('?')[1];
if(params.indexOf('&') > 2){
var vars = params.split('&');
for (var i in vars)
eval_var(vars[i].split('='));
}
else
eval_var(params.split('='));
}
return obj;
}
例子:
var url = "http://www.x.com?luckyNums=[31,21,6]&name=John&favFoods=[pizza]&noVal"
console.log(url2json(url));
输出:
[object]
noVal: ""
favFoods: "pizza"
name: "John"
luckyNums:
0: "31"
1: "21"
2: "6"