angularjs - Angular UI Bootstrap Popover 添加关闭按钮

Question

我有以下弹出窗口，并尝试添加一个关闭按钮以将其关闭。

.directive("popoverHtmlUnsafePopup", function () {
  'use strict';
  return {
    restrict: "EA",
    replace: true,
    scope: { title: "@", content: "@", placement: "@", animation: "&", isOpen: "&", manualHide: '&' },
    templateUrl: "views/popover/popover-html-unsafe-popup.html"
  };
})

.directive("popoverHtmlUnsafe", [ '$compile', '$timeout', '$parse', '$window',"$tooltip", function ($compile, $timeout, $parse, $window, $tooltip) {
  'use strict';
  return $tooltip("popoverHtmlUnsafe", "popover", "click");
}]);

<div class="popover {{placement}}" ng-class="{ in: isOpen(), fade: animation() }">
  <div class="arrow"></div>

  <div class="popover-inner">
    <h3 class="popover-title" ng-bind="title" ng-show="title"></h3>
    <div class="popover-content" bind-html-unsafe="content">
    </div>
    <button type="button" class="close" popover-trigger="close">&times;</button>
  </div>
</div>

只是不确定要调用什么事件或函数

<button type="button" class="close" popover-trigger="close">&times;</button>

能够关闭弹出窗口

Answer 1

在另一个使用工具提示的项目中遇到了这个问题。我最终遵循了 Tooltip.js 中的一些模式。

通过使用 $compile 和新的子作用域，您可以按照您认为合适的方式自定义此弹出窗口。

.directive('popover2',['$parse', '$compile','$log','$templateCache','$position',
    function ($parse,$compile,$log,$templateCache,$position ) {
         function link(scope, element, attrs) {
             var popupScope = scope.$new(false);   
             var html = $templateCache.get('views/popover/popover-html-unsafe-popup.html').trim();
             var template = angular.element(html)

             var popupLinkFn = $compile(template);

             var popup = popupLinkFn(popupScope);

             popupScope.isOpen = false;             
             popupScope.content = attrs['popover2'];
             var position = function(){
                 var ttPosition = $position.positionElements(element, popup, scope.placement, false);
                 ttPosition.top += 'px';
                 ttPosition.left +='px';
                 popup.css(ttPosition);
             };
             popupScope.setOpen = function(val) {                 

                 popupScope.isOpen = val;
                 popup.css({display: popupScope.isOpen ? 'block' :'none' });       
                 position();
                 // call twice, same as tooltip.js
                 position();

             };            

             var cleanup = element.on('click',function(event){   
                popupScope.setOpen(!popupScope.isOpen);
             });

             element.after(popup);
             element.on('$destroy',cleanup);

         }
        return {
            replace:false,            
            link:link,
            scope: {title: "@", placement: "@",},
        }; 
    }
 ]);

JSFiddle

该指令将允许您显示基于按钮的弹出窗口，然后具有关闭功能。您可以按照您认为合适的方式扩展显示逻辑。我过去也成功地使用过有效的表单。