如何让孩子不在 ag 网格中重复（角度）

Question

我试图让孩子应该只有 3 列基于assetCode 其中应该显示PRN、PRN1 和PRN2。

这是列表组件的代码

list.component.ts

this.rowData.push(
      {
        'code': 'Machine 1', 'assetCode': 'PRN', 'assetCount': 1,
        'date': '2019-01-18 00:00:00'
      },
      {
        'code': 'Machine 1', 'assetCode': 'PRN', 'assetCount': 1,
        'date': '2019-01-19 00:00:00'
      },
      {
        'code': 'Machine 2', 'assetCode': 'PRN 1', 'assetCount': 3,
        'date': '2019-01-20 00:00:00'
      },
      {
        'code': 'Machine 3', 'assetCode': 'PRN', 'assetCount': 1,
        'date': '2019-01-21 00:00:00'
      },
      {
        'code': 'Machine 4', 'assetCode': 'PRN 1', 'assetCount': 3,
        'date': '2019-01-22 00:00:00'
      },
      {
        'code': 'Machine 5', 'assetCode': 'PRN 2', 'assetCount': 3,
        'date': '2019-01-23 00:00:00'
      },
    );

    this.rowData = this.rowData.filter((item: any) => {
      return format(item.date, 'YYYY') === param;
    });

    const newData: any = [];
    this.rowData.forEach((x: any) => {
      const existing = newData.find((y: any) => format(y.date, 'YYYY-MM') === format(x.date, 'YYYY-MM')
        && y.assetCode === x.assetCode && y.code === x.code);
      if (existing) {
        existing.assetCount += existing.assetCount;
      } else {
        newData.push(x);
        this.rowData = newData;
      }
    });

    this.columnDefs.push(
      {
        'headerName': 'Style/Machine',
        'field': 'code',
        'pinned': 'left',
        'lockPosition': true
      }
    );

    for (let i = 0; i < 12; i++) {
      const record = {
        'headerName': this.monthNames[i].monthName,
        'children': [
          {
            'headerName': 'Total', 'columnGroupShow': 'closed',
            'valueGetter': ' ', 'headerClass': '', 'cellStyle': {}
          }
        ]
      };

      this.rowData.forEach((key: any) => {
        if (this.monthNames[i].monthName === format(key.date, 'MMMM')) {
          record.children.push(
            {
              'headerName': key.code, 'columnGroupShow': 'open', 'headerClass': 'header-text-center',
              'cellStyle': { 'justify-content': 'center' },
              'valueGetter': key.assetCount ? key.assetCount.toString() : null
            }
          );
        }
      });
      this.columnDefs.push(record);
    }
    }

输出

预期输出

提前感谢您的帮助

Answer 1

试试这个

const groupBy = (items, key) => items.reduce(
  (result, item) => ({
    ...result,
    [item[key]]: [
      ...(result[item[key]] || []),
      item,
    ],
  }), 
  {},
);
 var list= [{
        'code': 'Machine 1', 'assetCode': 'PRN', 'assetCount': 1,
        'date': '2019-01-18 00:00:00'
      },
      {
        'code': 'Machine 1', 'assetCode': 'PRN', 'assetCount': 1,
        'date': '2019-01-19 00:00:00'
      },
      {
        'code': 'Machine 2', 'assetCode': 'PRN 1', 'assetCount': 3,
        'date': '2019-01-20 00:00:00'
      },
      {
        'code': 'Machine 3', 'assetCode': 'PRN', 'assetCount': 1,
        'date': '2019-01-21 00:00:00'
      },
      {
        'code': 'Machine 4', 'assetCode': 'PRN 1', 'assetCount': 3,
        'date': '2019-01-22 00:00:00'
      },
      {
        'code': 'Machine 5', 'assetCode': 'PRN 2', 'assetCount': 3,
        'date': '2019-01-23 00:00:00'
      }];
      
const grouped = groupBy(list, 'code');
    
//console.log(grouped);
let newArr =[];
for (var key of Object.keys(grouped)) {
let obj ={};
obj["code"] =key;
grouped[key].forEach(function(element){
  //console.log(element);
  if (!obj.hasOwnProperty(element.assetCode)) 
  { 
    obj[element.assetCode] = 0;
  }
  obj[element.assetCode] = obj[element.assetCode] +element.assetCount;
});
newArr.push(obj);

    
}
console.log(newArr);