http://plnkr.co/edit/aiGCPJEwTj0RWnuoCjsW?p=preview
我希望删除按钮仅显示该项的弹出窗口。
你会怎么做? HTML:
<li ng-repeat="acc in accounts">
<div class="well well-sm">
<div class="popover-remove" ng-show="popoverRemove">Click to remove item</div>
<h4>{{acc.label}}</h4>
<button class="btn btn-default"
ng-mouseover="showPopover()"
ng-mouseleave="hidePopover()">Remove</button>
</div>
</li>
角度控制器:
var app = angular.module('myApp', [])
.controller('AccountsCtrl',
['$scope',
function($scope) {
var vs = $scope;
vs.accounts = [
{
id: '1',
label: 'Bitage'
},
{
id: '2',
label: 'Blockchain.info'
},
{
id: '3',
label: 'Coinbase wallet'
},
{
id: '4',
label: 'Xapo wallet'
}
];
vs.showPopover = function() {
vs.popoverRemove = true;
};
vs.hidePopover = function() {
vs.popoverRemove = false;
};
}]);
答案 0 :(得分:4)
Plunker为你
问题是ng-repeat创建了它自己的范围。因此,'popoverRemove'是每个范围的局部变量。您可以通过向ng-repeat的特定元素的控制器发送上下文来将true或false设置为局部变量并使用'this'设置它的值。
<button class="btn btn-default"
ng-mouseover="showPopover(this)"
ng-mouseleave="hidePopover(this)">Remove</button>
vs.showPopover = function(context) {
context.popoverRemove = true;
};
vs.hidePopover = function(context) {
context.popoverRemove = false;
};
答案 1 :(得分:2)
您还可以在每次迭代中创建一个属性,如下所示:
我没有在范围内调用函数,而只是将well.show
的值设置为true / false,但是你得到了这个想法的主旨!
<li ng-repeat="acc in accounts track by $index">
<div class="well well-sm">
<div class="popover-remove" ng-show="well.show">Click to remove item</div>
<h4>{{acc.label}}</h4>
<button class="btn btn-default"
ng-mouseover="well.show=true"
ng-mouseleave="well.show=false">Remove</button>
</div>
</li>
答案 2 :(得分:1)
将popOverRemove移动到每个帐户。然后,您可以从每个子范围控制它。同时更新你的showPopover / hidePopover。
vs.accounts = [
{
id: '1',
label: 'Bitage',
popoverRemove: false
},
{
id: '2',
label: 'Blockchain.info',
popoverRemove: false
},
{
id: '3',
label: 'Coinbase wallet',
popoverRemove: false
},
{
id: '4',
label: 'Xapo wallet',
popoverRemove: false
}
];
我已在此处更新