如何对数组对象进行分组以获得所需的数据格式？

Question

我尝试使用此对象数组对对象数据进行分组：

var data = [ 

    { IdUser: 8, Name: "John Smith", transferType: 'download', total: 6 },

    { IdUser: 12, Name: "Jane Smith", transferType: 'download', total: 15 },

    { IdUser: 11, Name: "Joe Smith", transferType: 'download', total: 20 },

    { IdUser: 12, Name: "Jane Smith", transferType: 'upload', total: 7 },

    { IdUser: 11, Name: "Joe Smith", transferType: 'upload', total: 16 },

    { IdUser: 8, Name: "John Smith", transferType: 'upload', total: 12 }

 ];

将所有对象的所需输出格式获取到新数组中：

  [

    {

       IdUser: 8,

       Name: 'John Smith',

       download: [{ IdUser: 8, Name: "John Smith", transferType: 'download', total: 6}],

       upload: [{ IdUser: 8, Name: "John Smith", transferType: 'upload', total: 12 }]

    },

    { 

       IdUser: 12,

       Name: 'Jane Smith',

       donwload: [{ IdUser: 12, Name: "Jane Smith", transferType: 'download', total: 15 }],

       upload: [{ IdUser: 12, Name: "Jane Smith", transferType: 'upload', total: 7 }]

    },

    {

       IdUser: 11,

       Name: 'Joe Smith',

       download: [{ IdUser: 11, Name: "Joe Smith", transferType: 'download', total: 20 }],

       upload: [{ IdUser: 11, Name: "Joe Smith", transferType: 'upload', total: 16 }]

    }

  ];

我尝试使用reduce函数，但它不是我得到的所需格式：

data.reduce(function (a, b) {

      a[b.Name] = a[b.Name] || [];   

      a[b.Name]["IdUser"] = b.IdUser;  

      a[b.Name][b.transferType] = a[b.Name][b.transferType] || [];     

      a[b.Name][b.transferType].push(b);

      return a;

    }, []);

Answer 1

我建议你分两次做。第一次使用分组功能按 IdUser 对数据进行分组。然后使用 for in 循环：var data = [  {    "IdUser": 8,    "Name": "John Smith",    "transferType": "download",    "total": 6  },  {    "IdUser": 12,    "Name": "Jane Smith",    "transferType": "download",    "total": 15  },  {    "IdUser": 11,    "Name": "Joe Smith",    "transferType": "downloaded",    "total": 20  },  {    "IdUser": 12,    "Name": "Jane Smith",    "transferType": "upload",    "total": 7  },  {    "IdUser": 11,    "Name": "Joe Smith",    "transferType": "upload",    "total": 16  },  {    "IdUser": 8,    "Name": "John Smith",    "transferType": "upload",    "total": 12  }];var groupBy = function(xs, key) {  return xs.reduce(function(rv, x) {    (rv[x[key]] = rv[x[key]] || []).push(x);    return rv;  }, {});};var tmp =  groupBy(data, 'IdUser');var formattedData = [];for(var id in tmp ){    formattedData.push({    "IdUser": id,    "Name": tmp[id][0].Name,    "download": tmp[id].filter(obj => obj.transferType == "download"),    "upload":tmp[id].filter(obj => obj.transferType == "upload")    })}console.log(formattedData)

Answer 2

编辑以使用reduce。var data = [{    IdUser: 8,    Name: "John Smith",    transferType: 'download',    total: 6  },  {    IdUser: 12,    Name: "Jane Smith",    transferType: 'download',    total: 15  },  {    IdUser: 11,    Name: "Joe Smith",    transferType: 'download',    total: 20  },  {    IdUser: 12,    Name: "Jane Smith",    transferType: 'upload',    total: 7  },  {    IdUser: 11,    Name: "Joe Smith",    transferType: 'upload',    total: 16  },  {    IdUser: 8,    Name: "John Smith",    transferType: 'upload',    total: 12  }];let ret = data.reduce((ret, cur) => {  let computed = ret.find(record => cur.IdUser === record.IdUser)  if (computed) {    computed[cur.transferType].push(cur)  } else {    let fresh = {      IdUser: cur.IdUser,      Name: cur.Name,      download: [],      upload: []    }    fresh[cur.transferType].push(cur)    ret.push(fresh)  }  return ret}, [])console.log(ret)

Answer 3

您可以使用该array.map方法。const myObj = [    {       IdUser: 8,       Name: 'John Smith',       download: [{ IdUser: 8, Name: "John Smith", transferType: 'download', total: 6}],       upload: [{ IdUser: 8, Name: "John Smith", transferType: 'upload', total: 12 }]    },    {        IdUser: 12,       Name: 'Jane Smith',       donwload: [{ IdUser: 12, Name: "Jane Smith", transferType: 'download', total: 15 }],       upload: [{ IdUser: 12, Name: "Jane Smith", transferType: 'upload', total: 7 }]    },    {       IdUser: 11,       Name: 'Joe Smith',       download: [{ IdUser: 11, Name: "Joe Smith", transferType: 'downloaded', total: 20 }],       upload: [{ IdUser: 11, Name: "Joe Smith", transferType: 'upload', total: 16 }]    }  ];const uploadData = myObject.map(val => val.upload);const downloadData = myObject.map(val => val.download);const newData = uploadData.concat(downloadData);

Answer 4

您可以使用一个简单的循环，例如：for&nbsp;(let&nbsp;i=0;&nbsp;i<data.length;&nbsp;i++)&nbsp;{&nbsp;...&nbsp;}在循环内部，您可以访问每个对象和属性，例如：data[i].IdUser