Question

当我在Javascript中处理对象数组时，我遇到了问题。

问题：输入


    [
    {  
        "artistName": "NgaNguyen Duy",
        "workId": "554a9e4fa36f794b31000024"
    }, 
    {
        "artistName": "NgaNguyen Duy",
        "workId": "554d3ffbcc477a7110000003"
    },
    {
        "artistName": "Kristin",
        "workId": "557a4kfbcc5e2a7110000223"
    }
    ]

输出应该是：


    [
    {
        "artistName": "NgaNguyen Duy",
        "workId"    : ["554a9e4fa36f794b31000024", "554d3ffbcc477a7110000003"]
    },
    {
        "artistName": "Kristin",
        "workId": ["557a4kfbcc5e2a7110000223"]
    }
    ]

提前感谢。

Answer 1

You can do aggregate :

var t =  [
    {  
        "artistName": "NgaNguyen Duy",
        "workId": "554a9e4fa36f794b31000024"
    }, 
    {
        "artistName": "NgaNguyen Duy",
        "workId": "554d3ffbcc477a7110000003"
    },
    {
        "artistName": "Kristin",
        "workId": "557a4kfbcc5e2a7110000223"
    }
    ];


var u = {};
var res = [];

t.forEach(function(item){
    if(!u[item.artistName]){        
         u[item.artistName] = [];  
    }
    u[item.artistName].push(item.workId);

});

for(var key in u){
    res.push({'artistName':key,'workId':u[key]});
}

console.log(res);


https://jsfiddle.net/Lgbcxorj/

Answer 2

所以你想合并“workId”的同名条目？：

var artists_in =     [
{  
    "artistName": "NgaNguyen Duy",
    "workId": "554a9e4fa36f794b31000024"
}, 
{
    "artistName": "NgaNguyen Duy",
    "workId": "554d3ffbcc477a7110000003"
},
{
    "artistName": "Kristin",
    "workId": "557a4kfbcc5e2a7110000223"
}
]

var artists_out = {}
for (i = 0; i < artists_in.length; i++)
{
  artistKey = artists_in[i]["artistName"];
  if(artists_out[artistKey] == null)
  {
    artists_out[artistKey] = { "artistName": artistKey, "workId": [artists_in[i]["workId"]] };
  }
  else
  {
    artists_out[artistKey]["workId"].push(artists_in[i]["workId"]);
  }
}

console.log(artists_out);

https://jsfiddle.net/hv9dtd8w/4/

在javascript中使用对象数组

2 个答案: