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Javascript 如何在我的案例中创建切换效果?_Javascript_Angularjs_Toggle - Fatal编程技术网

Javascript 如何在我的案例中创建切换效果?

Javascript 如何在我的案例中创建切换效果?,javascript,angularjs,toggle,Javascript,Angularjs,Toggle,我正在尝试在我的angular应用程序中创建一个切换 我有类似的东西 <div ng-repeat='item in items'> <div ng-click='toggle-{{item.number}} =!toggle-{{item.number}}> <span> <span>open</span> // I want to show 'open' text only when

我正在尝试在我的angular应用程序中创建一个切换

我有类似的东西

<div ng-repeat='item in items'>
    <div ng-click='toggle-{{item.number}} =!toggle-{{item.number}}>
        <span>
            <span>open</span>  // I want to show 'open' text only when the page first loaded
            <span>close</span> // I want to show 'open' text only when user clicks the div
        </span>
    </div>
</div>


你很接近!实际上,您甚至不需要在切换事件中包含
项.number
。您基本上希望翻转
showHide
值,并在两个选项中用
替换它们操作员

    <div ng-click="showHide = !showHide">
        <span>
            <span ng-show="!showHide">open</span> 
            <span ng-show="showHide">close</span>
        </span>
    </div>

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谢谢你的帮助。然而,我有ng repeat,它将创建许多具有open和class span的div。我一次只想打开/关闭一个div。如果我单击一个div,那么您的方法将打开所有span。+1考虑一下JSFIDLE,它只打开与该项目相关联的span,而不是所有span。(除非我很困惑lol)