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
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;
};
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.
}
可靠地做这件事比一开始想象的要复杂得多。
其他答案中使用的location.search很脆弱，应该避免使用-例如，如果有人搞砸了，并在？查询字符串。在我看来，URL在浏览器中自动转义的方式有很多种，这使得decodeURIComponent非常强制性。许多查询字符串是由用户输入生成的，这意味着对URL内容的假设非常糟糕。包括非常基本的东西，比如每个键都是唯一的，甚至有一个值。
为了解决这个问题，这里提供了一个可配置的API，并提供了健康的防御性编程。请注意，如果您愿意对某些变量进行硬编码，或者如果输入不能包含hasOwnProperty等，则可以将其大小减半。
版本1：返回包含每个参数的名称和值的数据对象。它有效地消除了重复，并始终尊重从左到右找到的第一个。
function getQueryData(url, paramKey, pairKey, missingValue, decode) {
var query, queryStart, fragStart, pairKeyStart, i, len, name, value, result;
if (!url || typeof url !== 'string') {
url = location.href; // more robust than location.search, which is flaky
}
if (!paramKey || typeof paramKey !== 'string') {
paramKey = '&';
}
if (!pairKey || typeof pairKey !== 'string') {
pairKey = '=';
}
// when you do not explicitly tell the API...
if (arguments.length < 5) {
// it will unescape parameter keys and values by default...
decode = true;
}
queryStart = url.indexOf('?');
if (queryStart >= 0) {
// grab everything after the very first ? question mark...
query = url.substring(queryStart + 1);
} else {
// assume the input is already parameter data...
query = url;
}
// remove fragment identifiers...
fragStart = query.indexOf('#');
if (fragStart >= 0) {
// remove everything after the first # hash mark...
query = query.substring(0, fragStart);
}
// make sure at this point we have enough material to do something useful...
if (query.indexOf(paramKey) >= 0 || query.indexOf(pairKey) >= 0) {
// we no longer need the whole query, so get the parameters...
query = query.split(paramKey);
result = {};
// loop through the parameters...
for (i = 0, len = query.length; i < len; i = i + 1) {
pairKeyStart = query[i].indexOf(pairKey);
if (pairKeyStart >= 0) {
name = query[i].substring(0, pairKeyStart);
} else {
name = query[i];
}
// only continue for non-empty names that we have not seen before...
if (name && !Object.prototype.hasOwnProperty.call(result, name)) {
if (decode) {
// unescape characters with special meaning like ? and #
name = decodeURIComponent(name);
}
if (pairKeyStart >= 0) {
value = query[i].substring(pairKeyStart + 1);
if (value) {
if (decode) {
value = decodeURIComponent(value);
}
} else {
value = missingValue;
}
} else {
value = missingValue;
}
result[name] = value;
}
}
return result;
}
}
版本2：返回一个具有两个相同长度数组的数据映射对象，一个用于名称，另一个用于值，每个参数都有一个索引。此格式支持重复名称，并故意不消除重复名称，因为这可能就是您希望使用此格式的原因。
function getQueryData(url, paramKey, pairKey, missingValue, decode) {
var query, queryStart, fragStart, pairKeyStart, i, len, name, value, result;
if (!url || typeof url !== 'string') {
url = location.href; // more robust than location.search, which is flaky
}
if (!paramKey || typeof paramKey !== 'string') {
paramKey = '&';
}
if (!pairKey || typeof pairKey !== 'string') {
pairKey = '=';
}
// when you do not explicitly tell the API...
if (arguments.length < 5) {
// it will unescape parameter keys and values by default...
decode = true;
}
queryStart = url.indexOf('?');
if (queryStart >= 0) {
// grab everything after the very first ? question mark...
query = url.substring(queryStart + 1);
} else {
// assume the input is already parameter data...
query = url;
}
// remove fragment identifiers...
fragStart = query.indexOf('#');
if (fragStart >= 0) {
// remove everything after the first # hash mark...
query = query.substring(0, fragStart);
}
// make sure at this point we have enough material to do something useful...
if (query.indexOf(paramKey) >= 0 || query.indexOf(pairKey) >= 0) {
// we no longer need the whole query, so get the parameters...
query = query.split(paramKey);
result = {
names: [],
values: []
};
// loop through the parameters...
for (i = 0, len = query.length; i < len; i = i + 1) {
pairKeyStart = query[i].indexOf(pairKey);
if (pairKeyStart >= 0) {
name = query[i].substring(0, pairKeyStart);
} else {
name = query[i];
}
// only continue for non-empty names...
if (name) {
if (decode) {
// unescape characters with special meaning like ? and #
name = decodeURIComponent(name);
}
if (pairKeyStart >= 0) {
value = query[i].substring(pairKeyStart + 1);
if (value) {
if (decode) {
value = decodeURIComponent(value);
}
} else {
value = missingValue;
}
} else {
value = missingValue;
}
result.names.push(name);
result.values.push(value);
}
}
return result;
}
}
// Parse query string
var params = {}, queryString = location.hash.substring(1),
regex = /([^&=]+)=([^&]*)/g,
m;
while (m = regex.exec(queryString)) {
params[decodeURIComponent(m[1])] = decodeURIComponent(m[2]);
}
这会奏效的。。。您需要在需要通过传递其名称来获取参数的地方调用此函数。。。
function getParameterByName(name)
{
name = name.replace(/[\[]/,"\\\[").replace(/[\]]/,"\\\]");
var regexS = "[\\?&]"+name+"=([^&#]*)";
var regex = new RegExp( regexS );
var results = regex.exec( window.location.href );
alert(results[1]);
if (results == null)
return "";
else
return results[1];
}
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
