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
snipplr.com上的Roshambo有一个简单的脚本来实现这一点,这在使用jQuery|Improved获取URL参数中有所描述。使用他的脚本,您还可以轻松地提取所需的参数。
要点如下:
$.urlParam = function(name, url) {
if (!url) {
url = window.location.href;
}
var results = new RegExp('[\\?&]' + name + '=([^&#]*)').exec(url);
if (!results) {
return undefined;
}
return results[1] || undefined;
}
然后从查询字符串中获取参数。
那么,如果URL/查询字符串是xyz.example/index.html?lang=de。
只需调用var langval=$.urlParam('lang');,你已经得到了。
UZBEKJON也在这方面发表了一篇很棒的博客文章,用jQuery获取URL参数和值。
其他回答
我使用以下代码(JavaScript)获取通过URL传递的内容:
function getUrlVars() {
var vars = {};
var parts = window.location.href.replace(/[?&]+([^=&]+)=([^&]*)/gi, function(m,key,value) {
vars[key] = value;
});
return vars;
}
然后,要将值分配给变量,只需指定要获取的参数,例如,如果URL是example.com/?I=1&p=2&f=3
您可以执行此操作以获取值:
var getI = getUrlVars()["I"];
var getP = getUrlVars()["p"];
var getF = getUrlVars()["f"];
则值将为:
getI = 1, getP = 2 and getF = 3
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;
};
非常轻量级的jQuery方法:
var qs = window.location.search.replace('?','').split('&'),
request = {};
$.each(qs, function(i,v) {
var initial, pair = v.split('=');
if(initial = request[pair[0]]){
if(!$.isArray(initial)) {
request[pair[0]] = [initial]
}
request[pair[0]].push(pair[1]);
} else {
request[pair[0]] = pair[1];
}
return;
});
console.log(request);
例如,提醒?q
alert(request.q)
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;
};
我经常使用正则表达式,但并非如此。
在我的应用程序中读取一次查询字符串,并从所有键/值对构建一个对象,对我来说似乎更容易、更有效,比如:
var search = function() {
var s = window.location.search.substr(1),
p = s.split(/\&/), l = p.length, kv, r = {};
if (l === 0) {return false;}
while (l--) {
kv = p[l].split(/\=/);
r[kv[0]] = decodeURIComponent(kv[1] || '') || true;
}
return r;
}();
对于URL,如http://example.com?param1=val1¶m2=val2您可以稍后在代码中以search.param1和search.param2的形式获取它们的值。