mog*_*gli 4 javascript typescript angular
您好,我想切换一个特定面板,但如果我单击该按钮,其他对象的面板将打开。我怎样才能打开点击的面板?
切换组件.ts
opened:Boolean=false;
toggle () {
this.opened = !this.opened;
}
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超文本标记语言
<div class="main" *ngFor="let x of data; let i=index;">
<footer>
<div class="icons">
<span id="{{item.id}}" (click)="toggle()">6<i class="fa fa-users {{i}}" ></i></span>
<span >6<i class="glyphicon glyphicon-picture"></i></span>
<span >6<i class="glyphicon glyphicon-tag"></i></span>
<div class="iconsRight pull-right">
<span >EXIF<i class="glyphicon glyphicon-info-sign"></i></span>
<span ><i class="fa fa-map-marker"></i></span>
<span ><i class="fa fa-share-alt-square"></i></span>
</div>
</div>
</footer>
<div class="togglePanel{{item.id}}" *ngIf="opened" >
<hr/>
<ul class="toggleWrapper">
<li>YES</li>
<hr/>
<li>YES</li>
<hr/>
<li>YES</li>
<hr/>
<li>YES</li>
</ul>
</div>
</div>
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您需要保存各个面板的状态。目前,您只需设置一个变量来切换所有面板。
在您的 中toggle.component.ts,添加一个变量来存储每个项目的状态:
togglePanel: any = {};
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然后,将您的 html 更改为以下内容:
<div class="main" *ngFor="let x of data; let i=index;">
<footer>
<div class="commentAgent">Text des Bewerters, der die Bearbeitung dieses Bildes vorgenommen hat</div>
<div class="icons">
<span id="{{item.id}}" (click)="togglePanel[i] = !togglePanel[i]">6<i class="fa fa-users {{i}}" ></i></span>
<span>6<i class="glyphicon glyphicon-picture"></i></span>
<span>6<i class="glyphicon glyphicon-tag"></i></span>
<div class="iconsRight pull-right">
<span>EXIF<i class="glyphicon glyphicon-info-sign"></i>/span>
<span><i class="fa fa-map-marker"></i></span>
<span><i class="fa fa-share-alt-square"></i></span>
</div>
</div>
</footer>
<div class="togglePanel{{item.id}}" *ngIf="togglePanel[i]">
<hr/>
<ul class="toggleWrapper">
<li>YES</li>
<hr/>
<li>YES</li>
<hr/>
<li>YES</li>
<hr/>
<li>YES</li>
</ul>
</div>
</div>
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此外,您不需要此方法中的toggle()方法和变量opened。
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