我有一个如下所述的对象数组,
var arraydata =[
{ id:1, name:"Abraham", age:20,gender:"male"},
{ id:2, name:"Annie", age:25,gender:"female"},
{ id:3, name:"Ryan", age:40,gender:"male"},
{ id:4, name:"Wayne", age:31,gender:"male"},
{ id:5, name:"Paul", age:45,gender:"male"}
];
如何将其转换为只有“id”和“name”字段的JSON数据?
答案 0 :(得分:3)
语法错误age=20,gender="male"
将其更改为age:20,gender:"male"
var arraydata =[
{ id:1, name:"Abraham", age: 20, gender:"male"},
{ id:2, name:"Annie", age: 25, gender:"female"},
{ id:3, name:"Ryan", age: 28, gender:"male"},
{ id:4, name:"Wayne", age: 19, gender:"male"},
{ id:5, name:"Paul", age: 45,gender:"male"}
];
var ans= arraydata.map(function(a){
return {id:a.id,name:a.name};
})
console.log(ans);
答案 1 :(得分:0)
var arraydata =[
{ id:1, name:"Abraham", age: 20, gender: "male"},
{ id:2, name:"Annie", age: 25, gender: "female"},
{ id:3, name:"Ryan", age: 28, gender: "male"},
{ id:4, name:"Wayne", age: 19, gender: "male"},
{ id:5, name:"Paul", age: 45, gender: "male"}
];
var _ = arraydata.map(function(item) {
return { id: item.id, name: item.name };
});
var jsonData = JSON.stringify(_);
console.log(jsonData);
您当然可以将这两个步骤链接在一起。
答案 2 :(得分:0)
您可以使用从对象中删除属性的delete operator
。
var arraydata = [{
id: 1,
name: "Abraham",
age : 20,
gender : "male"
}, {
id: 2,
name: "Annie",
age : 25,
gender : "female"
}, {
id: 3,
name: "Ryan",
age : 40,
gender : "male"
}, {
id: 4,
name: "Wayne",
age : 31,
gender : "male"
}, {
id: 5,
name: "Paul",
age : 45,
gender : "male"
}];
arraydata.forEach(function(item){ delete item.age; delete item.gender });
console.log(arraydata);