Javascript _co.open不是函数：基于条件的运行方法

我有一个方法，这个方法应该根据条件执行，在这里我打开一个模式窗口，点击地图上的标记

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

将open方法封装在if语句“TypeError:\u co.open不是函数”中后，会出现此错误。要根据条件打开模式窗口

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

将条件放在打开的内部：

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