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Angular 如何根据其他表单字段值禁用离子表单字段?_Angular_Ionic Framework - Fatal编程技术网

Angular 如何根据其他表单字段值禁用离子表单字段?

Angular 如何根据其他表单字段值禁用离子表单字段?,angular,ionic-framework,Angular,Ionic Framework,有一种离子选择 <ion-select interface="popover" formControlName="window_exists" (ionChange)="openWindow()" type="text"> <ion-option value="Y">Yes</ion-option> <ion-option value="N">No</ion-option> &

有一种离子选择

 <ion-select interface="popover" formControlName="window_exists"  (ionChange)="openWindow()" type="text">
          <ion-option value="Y">Yes</ion-option>
          <ion-option value="N">No</ion-option>
        </ion-select>

对
不
如果离子选项值为“否”,我想禁用离子输入



请引导

您需要将$event作为参数传递

<ion-select interface="popover" formControlName="window_exists"  (ionChange)="openWindow($event)" type="text">
  <ion-option value="Y">Yes</ion-option>
  <ion-option value="N">No</ion-option>
</ion-select>

谢谢。

再次感谢AddWeb:)你能检查一下这个问题吗
<ion-select interface="popover" formControlName="window_exists"  (ionChange)="openWindow($event)" type="text">
  <ion-option value="Y">Yes</ion-option>
  <ion-option value="N">No</ion-option>
</ion-select>
<ion-input disabled="{{isDisabled}}" type="number" min="0" formControlName="ml_window_no" required clearInput></ion-input>
isDisabled:boolean = false;

openWindow(event){
     console.log('event',event);
     if (event == 'N') {
       console.log('no');
       this.isDisabled = true;
     }else{
       console.log('yes');
       this.isDisabled = false;
     }
}