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Angular @子组件的输出不';t触发事件发射器角度5_Angular_Eventemitter - Fatal编程技术网

Angular @子组件的输出不';t触发事件发射器角度5

Angular @子组件的输出不';t触发事件发射器角度5,angular,eventemitter,Angular,Eventemitter,我试图使用@Output和事件发射器在子组件和父组件之间进行通信,以传递变量。我遵循了本教程: 我已经研究了与此相关的不同问题,但都不适用于我的案例 子HTML <button (click)="onLinkedClicked();">Click me</button> <showcase-menu #topNavMenu [showBranding]="false" [showSearch]= "false" (onLinkedClicked)="l

我试图使用@Output和事件发射器在子组件和父组件之间进行通信,以传递变量。我遵循了本教程:

我已经研究了与此相关的不同问题,但都不适用于我的案例

子HTML

<button (click)="onLinkedClicked();">Click me</button>
 <showcase-menu #topNavMenu [showBranding]="false"
     [showSearch]= "false" (onLinkedClicked)="linkToggler($event)"></showcase-menu>

linkToggler()永远不会被调用。我在控制台中没有任何错误,

事件必须与与输出装饰器关联的属性名称匹配

<showcase-menu #topNavMenu 
               [showBranding]="false"
               [showSearch]= "false"  
               (linkedClicked)="linkToggler($event)">. 
</showcase-menu>

事件必须与输出装饰器关联的属性名称匹配

<showcase-menu #topNavMenu 
               [showBranding]="false"
               [showSearch]= "false"  
               (linkedClicked)="linkToggler($event)">. 
</showcase-menu>
<showcase-menu #topNavMenu 
               [showBranding]="false"
               [showSearch]= "false"  
               (linkedClicked)="linkToggler($event)">. 
</showcase-menu>