【问题标题】:Angular UI Bootstrap Popover adding a close buttonAngular UI Bootstrap Popover 添加关闭按钮
【发布时间】:2015-04-02 14:25:54
【问题描述】:

我有以下弹出窗口并尝试添加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

【问题讨论】:

    标签: angularjs angular-ui-bootstrap popover


    【解决方案1】:

    在使用工具提示的另一个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 和一个新的子范围,您可以自定义此弹出窗口,以您认为合适的方式。

    .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。我过去也成功使用过有效的表单。

    【讨论】:

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