是否有一个快速和简单的方法来编码JavaScript对象到字符串,我可以通过GET请求传递?
没有jQuery,没有其他框架-只有纯JavaScript:)
是否有一个快速和简单的方法来编码JavaScript对象到字符串,我可以通过GET请求传递?
没有jQuery,没有其他框架-只有纯JavaScript:)
当前回答
我有一个更简单的解决方案,它不使用任何第三方库,并且已经易于在任何具有“Object”的浏览器中使用。键”(也就是所有现代浏览器+ edge + ie):
In ES5
function(a){
if( typeof(a) !== 'object' )
return '';
return `?${Object.keys(a).map(k=>`${k}=${a[k]}`).join('&')}`;
}
在胡
function(a){
if( typeof(a) !== 'object' )
return '';
return '?' + Object.keys(a).map(function(k){ return k + '=' + a[k] }).join('&');
}
其他回答
如果您想递归地转换嵌套对象,并且对象可能包含也可能不包含数组(数组可能包含对象或数组等),那么解决方案就会变得稍微复杂一些。这就是我的尝试。
我还添加了一些选项,用于选择是否要为每个对象成员记录它位于主对象中的深度,以及选择是否要向来自已转换数组的成员添加标签。
理想情况下,您应该测试thing参数是否真的接收到对象或数组。
function thingToString(thing,maxDepth,recordLevel,markArrays){
//thing: object or array to be recursively serialized
//maxDepth (int or false):
// (int) how deep to go with converting objects/arrays within objs/arrays
// (false) no limit to recursive objects/arrays within objects/arrays
//recordLevel (boolean):
// true - insert "(level 1)" before transcript of members at level one (etc)
// false - just
//markArrays (boolean):
// insert text to indicate any members that came from arrays
var result = "";
if (maxDepth !== false && typeof maxDepth != 'number') {maxDepth = 3;}
var runningDepth = 0;//Keeps track how deep we're into recursion
//First prepare the function, so that it can call itself recursively
function serializeAnything(thing){
//Set path-finder values
runningDepth += 1;
if(recordLevel){result += "(level " + runningDepth + ")";}
//First convert any arrays to object so they can be processed
if (thing instanceof Array){
var realObj = {};var key;
if (markArrays) {realObj['type'] = "converted array";}
for (var i = 0;i < thing.length;i++){
if (markArrays) {key = "a" + i;} else {key = i;}
realObj[key] = thing[i];
}
thing = realObj;
console.log('converted one array to ' + typeof realObj);
console.log(thing);
}
//Then deal with it
for (var member in thing){
if (typeof thing[member] == 'object' && runningDepth < maxDepth){
serializeAnything(thing[member]);
//When a sub-object/array is serialized, it will add one to
//running depth. But when we continue to this object/array's
//next sibling, the level must go back up by one
runningDepth -= 1;
} else if (maxDepth !== false && runningDepth >= maxDepth) {
console.log('Reached bottom');
} else
if (
typeof thing[member] == "string" ||
typeof thing[member] == 'boolean' ||
typeof thing[member] == 'number'
){
result += "(" + member + ": " + thing[member] + ") ";
} else {
result += "(" + member + ": [" + typeof thing[member] + " not supported]) ";
}
}
}
//Actually kick off the serialization
serializeAnything(thing);
return result;
}
以下是已接受答案的CoffeeScript版本。
serialize = (obj, prefix) ->
str = []
for p, v of obj
k = if prefix then prefix + "[" + p + "]" else p
if typeof v == "object"
str.push(serialize(v, k))
else
str.push(encodeURIComponent(k) + "=" + encodeURIComponent(v))
str.join("&")
Ramda:
R.pipe(R.toPairs, R.map(R.join('=')), R.join('&'))({a: 'b', b: 'a'})
这里有一个简单的答案,在转换过程中同时处理字符串和数组。
jsonToQueryString: function (data) {
return Object.keys(data).map((key) => {
if (Array.isArray(data[key])) {
return (`${encodeURIComponent(key)}=${data[key].map((item) => encodeURIComponent(item)).join('%2C')}`);
}
return(`${encodeURIComponent(key)}=${encodeURIComponent(data[key])}`);
}).join('&');
}
是这样的:
Serialize =函数(obj) { Var STR = []; For (var p in obj) if (obj.hasOwnProperty(p)) { str.push(encodeURIComponent(p) + "=" + encodeURIComponent(obj[p]))); } 返回str.join(“&”); } console.log(序列化({ Foo:“你好”, 栏:“100%” })); / / foo =你好% 20 there&bar 25 = 100%
它还转换递归对象(使用PHP“数组”符号表示查询字符串):
serialize = function(obj, prefix) { var str = [], p; for (p in obj) { if (obj.hasOwnProperty(p)) { var k = prefix ? prefix + "[" + p + "]" : p, v = obj[p]; str.push((v !== null && typeof v === "object") ? serialize(v, k) : encodeURIComponent(k) + "=" + encodeURIComponent(v)); } } return str.join("&"); } console.log(serialize({ foo: "hi there", bar: { blah: 123, quux: [1, 2, 3] } })); // foo=hi%20there&bar%5Bblah%5D=123&bar%5Bquux%5D%5B0%5D=1&bar%5Bquux%5D%5B1%5D=2&bar%5Bquux%5D%5B2%5D=3