如何将JavaScript对象转换为字符串?
例子:
var o = {a:1, b:2}
console.log(o)
console.log('Item: ' + o)
输出:
对象{a=1, b=2} //非常好的可读输出:) Item: [object object] //不知道里面有什么:(
如何将JavaScript对象转换为字符串?
例子:
var o = {a:1, b:2}
console.log(o)
console.log('Item: ' + o)
输出:
对象{a=1, b=2} //非常好的可读输出:) Item: [object object] //不知道里面有什么:(
当前回答
setobjToString:function(obj){
var me =this;
obj=obj[0];
var tabjson=[];
for (var p in obj) {
if (obj.hasOwnProperty(p)) {
if (obj[p] instanceof Array){
tabjson.push('"'+p +'"'+ ':' + me.setobjToString(obj[p]));
}else{
tabjson.push('"'+p +'"'+':"'+obj[p]+'"');
}
}
} tabjson.push()
return '{'+tabjson.join(',')+'}';
}
其他回答
如果你不播放join()到对象。
const obj = {one:1, two:2, three:3};
let arr = [];
for(let p in obj)
arr.push(obj[p]);
const str = arr.join(',');
使用javascript String()函数
String(yourobject); //returns [object Object]
或stringify ()
JSON.stringify(yourobject)
JSON似乎接受了第二个参数,可以帮助函数- replace,这以最优雅的方式解决了转换问题:
JSON.stringify(object, (key, val) => {
if (typeof val === 'function') {
return String(val);
}
return val;
});
循环引用
通过使用下面的替换器,我们可以产生更少冗余的JSON -如果源对象包含对某个对象的多次引用,或者包含循环引用-那么我们通过特殊的路径字符串(类似于JSONPath)引用它-我们如下所示
let s = JSON.stringify(obj, refReplacer());
function refReplacer() { let m = new Map(), v= new Map(), init = null; return function(field, value) { let p= m.get(this) + (Array.isArray(this) ? `[${field}]` : '.' + field); let isComplex= value===Object(value) if (isComplex) m.set(value, p); let pp = v.get(value)||''; let path = p.replace(/undefined\.\.?/,''); let val = pp ? `#REF:${pp[0]=='[' ? '$':'$.'}${pp}` : value; !init ? (init=value) : (val===init ? val="#REF:$" : 0); if(!pp && isComplex) v.set(value, path); return val; } } // --------------- // TEST // --------------- // gen obj with duplicate references let a = { a1: 1, a2: 2 }; let b = { b1: 3, b2: "4" }; let obj = { o1: { o2: a }, b, a }; // duplicate reference a.a3 = [1,2,b]; // circular reference b.b3 = a; // circular reference let s = JSON.stringify(obj, refReplacer(), 4); console.log(s);
奖励:这里是这种序列化的逆函数
function parseRefJSON(json) { let objToPath = new Map(); let pathToObj = new Map(); let o = JSON.parse(json); let traverse = (parent, field) => { let obj = parent; let path = '#REF:$'; if (field !== undefined) { obj = parent[field]; path = objToPath.get(parent) + (Array.isArray(parent) ? `[${field}]` : `${field?'.'+field:''}`); } objToPath.set(obj, path); pathToObj.set(path, obj); let ref = pathToObj.get(obj); if (ref) parent[field] = ref; for (let f in obj) if (obj === Object(obj)) traverse(obj, f); } traverse(o); return o; } // ------------ // TEST // ------------ let s = `{ "o1": { "o2": { "a1": 1, "a2": 2, "a3": [ 1, 2, { "b1": 3, "b2": "4", "b3": "#REF:$.o1.o2" } ] } }, "b": "#REF:$.o1.o2.a3[2]", "a": "#REF:$.o1.o2" }`; console.log('Open Chrome console to see nested fields:'); let obj = parseRefJSON(s); console.log(obj);
如果对象是一个jQuery对象,那么你应该使用:
obj.html()
而不是:
JSON.stringify(obj)
例子:
Var tr = $('tr') console.log('This does not work:') console.log (JSON.stringify (tr)) console.log('But this does:') console.log (tr.html ()) < script src = " https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js " > < /脚本> <表> < tr > < td > < / td > < td > b < / td > 表> < /