Question

如何在函数onSelectedReport()中将两个给定条件合并为一个。如果用html编写这些条件，它将看起来像这样：

html：

  <div *ngFor="let report of reports">
    <div *ngFor="let i  of income">
      <div *ngIf="report.r_income_id == i.income_id">
        <div *ngFor="let c  of costs">
          <div *ngIf="i.i_costs_id == c.costs_id">
            {{c.name}}
          </div>
        </div>
       </div>
    </div>  
  </div>

但是由于我需要在选定的特定标识符上显示数据，因此需要在ts上实现它。我尚未对其进行测试，但是我知道100％它将无法正常工作，因为在第二种情况下this.income.i_costs_id的值将确定undefined。并且这两个条件很可能应该合并为一个。该怎么办？

ts：

  reports: Reports[]
  income: Income[]
  costs: Costs[]
  selectedReport = null
  filteredIncome = []
  filteredСosts = []

  onSelectedReport(reportId) {
    this.selectedReport = this.reports.find(
      el => {
        return el.report_id === reportId
      }
    )
    if (this.incomeService) {
      this.incomeService.fetchAll().subscribe(
        income => {
          this.income = income
          this.filteredIncome = this.income.filter(
            (income) => income.income_id == this.selectedReport.r_income_id
          )
        }
      )
    }
    if (this.costsService) {
      this.costsService.fetch().subscribe(
        costs => {
          this.costs = costs
          this.filteredСosts = this.costs.filter(
            (costs) => costs.costs_id == this.income.i_costs_id
          )
        }
      )
    }
  }

Answer 1

尝试

  reports: Reports[]
  income: Income[]
  costs: Costs[]
  selectedReport = null
  filteredIncome = []
  filteredСosts = []

  onSelectedReport(reportId) {
    this.selectedReport = this.reports.find(
      el => {
        return el.report_id === reportId
      }
    )
    if (this.incomeService) {
      this.incomeService.fetchAll().subscribe(
        income => {
          this.income = income
          this.filteredIncome = this.income.filter(
            (income) => income.income_id == this.selectedReport.r_income_id
          )
          if (this.costsService) {
            this.costsService.fetch().subscribe(
              costs => {
                this.costs = costs
                for(let i of this.filteredIncome){
                  for(let c of costs){
                    if(c.costs_id==i.i_costs_id){
                      this.filteredСosts.push(c)
                    }
                  }
                }
              }
            )
          }
        }
      )
    }
  }

Answer 2

您的要求符合forkJoin，我在此处添加了示例，您可以根据您的代码对其进行修改。

let requests = [];

if (this.incomeService) {
  //adding income service
  requests.push(this.incomeService.fetchAll());
} else {
  //adding null as we must need observable of first one
  requests.push(Observable.of(null))
}

if (this.costsService) {
  //adding cost service
  requests.push(this.costsService.fetch());
} else {
  //adding null of cost
  requests.push(Observable.of(null))
}

//requesting the requests
Observable.forkJoin(...requests)
  .subscribe(result => {
    //accessing the values
    console.log(values[0], values[1]);
    //check if values[0] is not null first so second one does not call too
    //here values[0] is first one and values[1] is costs service call
  });

如何通过选定的值过滤数据？

2 个答案: