我有这个复杂的对象结构:
myObject = {
"myObject" : [
{
"id" : 1,
"parameters" : [
{
"name" : "name1",
"special" : "xxx"
},
{
"name" : "name2",
"special" : "yyy"
}
]
},
{
"id" : 2,
"parameters" : [
{
"name" : "name3",
"special" : "zzz"
}
]
},
{
"id" : 2,
"parameters" : [
{
"name" : "name4",
"special" : "ttt"
},
{
"name" : "name5",
"special" : "aaa"
},
{
"name" : "name6",
"special" : "zzz"
}
]
},
...
]
};
它由一系列其他对象组成,每个对象的变量都为parameters
。
我的目标是将每个对象的special
parameters
连接成一个新字符串,该字符串必须存储为新属性。
在这种情况下,结果应如下所示:
myObject = {
"myObject" : [
{
"id" : 1,
"parameters" : [
{
"name" : "name1",
"special" : "xxx"
},
{
"name" : "name2",
"special" : "yyy"
}
],
"newProp" : "xxxyyy"
},
{
"id" : 2,
"parameters" : [
{
"name" : "name3",
"special" : "zzz"
}
],
"newProp" : "zzz"
},
{
"id" : 2,
"parameters" : [
{
"name" : "name4",
"special" : "ttt"
},
{
"name" : "name5",
"special" : "aaa"
},
{
"name" : "name6",
"special" : "zzz"
}
],
"newProp" : "tttaaazzz"
},
...
]
};
我试过这样的事情:
forEach(arr in myObject.myObject){
arr.parameters(forEach (i in arr.parameters.special) {
myObject.myObject = i.concat(myObject.myObject);
})
}
显然,它不起作用。但我想这可能是正确的做法。
有什么建议吗?
答案 0 :(得分:5)
您可以使用Array#forEach
遍历对象,然后使用Array#map
和Array#join
根据参数值构建字符串,如下所示:
const myObject = {"myObject":[{"id":1,"parameters":[{"name":"name1","special":"xxx"},{"name":"name2","special":"yyy"}]},{"id":2,"parameters":[{"name":"name3","special":"zzz"}]},{"id":2,"parameters":[{"name":"name4","special":"ttt"},{"name":"name5","special":"aaa"},{"name":"name6","special":"zzz"}]}]};
myObject.myObject.forEach(item => {
item.newProp = item.parameters.map(p => p.special).join('');
});
console.log(myObject);

答案 1 :(得分:1)
使用reduce和for Each
var myObject = {
"myObject" : [
{
"id" : 1,
"parameters" : [
{
"name" : "name1",
"special" : "xxx"
},
{
"name" : "name2",
"special" : "yyy"
}
]
},
{
"id" : 2,
"parameters" : [
{
"name" : "name3",
"special" : "zzz"
}
]
},
{
"id" : 2,
"parameters" : [
{
"name" : "name4",
"special" : "ttt"
},
{
"name" : "name5",
"special" : "aaa"
},
{
"name" : "name6",
"special" : "zzz"
}
]
}
]
};
myObject.myObject.forEach(arr => {
arr.prop = arr.parameters.reduce((res,obj)=> res+obj.special, '')
})
console.log(myObject)

答案 2 :(得分:1)
您可以像这样使用.map()
和.reduce()
:
let myObject = [{"id" : 1, "parameters" : [{ "name" : "name1", "special" : "xxx"}, { "name" : "name2", "special" : "yyy" }]}, { "id" : 2, "parameters" : [{ "name" : "name3", "special" : "zzz"}]}, {"id" : 2, "parameters" : [{ "name" : "name4", "special" : "ttt"}, { "name" : "name5", "special" : "aaa"},{ "name" : "name6", "special" : "zzz"}]}];
let result = myObject.map(
o => (o.newProp = o['parameters'].reduce((a, o) => a + o['special'], ""), o)
);
console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

答案 3 :(得分:1)
另一种方法是使用嵌套的地图函数:
myObject = {"myObject":[{"id":1,"parameters":[{"name":"name1","special":"xxx"},{"name":"name2","special":"yyy"}]},{"id":2,"parameters":[{"name":"name3","special":"zzz"}]},{"id":2,"parameters":[{"name":"name4","special":"ttt"},{"name":"name5","special":"aaa"},{"name":"name6","special":"zzz"}]}]};
myObject.myObject.map(x => {
x.newProp = x.parameters.map(p => p.special).join('');
return x;
})
console.log(myObject);