javascript - _co.open 不是函数 :run method based on condition

Question

我有一个方法，该方法应该根据条件执行，这里我在 map 上点击标记打开一个模式窗口。

    if(this.showMobile){
open(content) {
  this.modalService.open(content).result.then((result) => {
  this.closeResult = `Closed with: ${result}`;
  }, (reason) => {
  this.closeResult = `Dismissed ${this.getDismissReason(reason)}`;
  });
  }
}

在 if 语句“TypeError: _co.open is not a function”中包含该 open 方法后出现此错误。想根据条件打开模态窗口。

<div id="dvMap"  (click)="open(content)"  style="width:900px;height:600px;"></div>

Answer 1

将条件置于内部打开:

open(content) {
  if(this.showMobile) {
    this.modalService.open(content).result.then((result) => {
      this.closeResult = `Closed with: ${result}`;
    }, (reason) => {
      this.closeResult = `Dismissed ${this.getDismissReason(reason)}`;
    });
  }
}