Question

我遇到了如何在数组中循环和存储的问题。

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对于/** Model to display response*/ export class ResultsData { id: number, name: string, startDate: string, endDarte: string, startTime: string, endTime: string, city: string, country: string } /**Response upon service call*/ response = [ { startDate: '04/05/2018', endDate: '05/05/2018' , personDetails : [ { id: 5, name: 'kumar', timeDetails: [{ startTime: ’09:22pm’, endTime: ’08:33’}, { startTime: ’01:22pm’, endTime: ’01:33’}] }, { id: 4, name: 'vishal', timeDetails: [{ startTime: ’09:22pm’, endTime: ’08:33’}, { startTime: ’01:22pm’, endTime: ’01:33’}] } , { id: 2, name: 'dinesh', timeDetails: [{ startTime: ’09:22pm’, endTime: ’08:33’}, { startTime: ’01:22pm’, endTime: ’01:33’}] }], locationDetails : [{city: 'new york', country: 'us'}, {city: 'los angeles', country: 'us'} ]} ] /**Component.ts*/ tableData : Array<ResultsData> = []中的每个人，我想将他的详细信息（EG。名称和ID）和时间详细信息的最后一个索引，位置详细信息的最后一个索引推送到数组中。

personDetails

Answer 1

我建议您尝试使用构造函数方法将personDetails映射到tableData：

class ResultsData {
    id: number;
    name: string;
    startDate: string;
    endDate: string;
    startTime: string;
    endTime: string;
    city: string;
    country: string;

    constructor(item) {
      Object.assign(this, item);
    }
}

tableData = response[0].personDetails.map(item => new ResultsData ({
  id: Number(item.id),
  name: item.name,
  startDate: response[0].startDate,
  endDate: response[0].endDate,
  startTime: item.timeDetails[item.timeDetails.length - 1].startTime,
  endTime: item.timeDetails[item.timeDetails.length - 1].endTime,
  city: response[0].locationDetails[response[0].locationDetails.length - 1].city,
  country: response[0].locationDetails[response[0].locationDetails.length - 1].country
}));

循环响应的功能

1 个答案: