Question

我正在使用ui-bootstrap分页，根据点击的页面从服务器获取数据。每次它都会像这样将请求发送到服务器。＆＃39; apiURL？skip =＆＃39; +0，如果我点击后退按钮，它将更新网址，而不是视图，页码也会在上一页保持活动状态。而且当我直接在浏览器中放置一些页码时，我想要工作。任何人都可以帮我如何处理页面网址。

mainEvents: function() {
    var self = this;
    var currentItem =0;
    return $http.get(UrlService.baseUrl + '/event/upcoming?skip='+ currentItem ).then(function(response) {
        var events = response.data.events;
        var eventsLen = events.length;
        for (var i = 0; i < eventsLen; i++) {
            self.prepareForRendering(events[i]);
        }
        var totalItems = response.data.events.length;
        angular.copy(response.data.events, upEvents);
        return events;
    }, function(response) {
        return $q.reject(response.data.error)
    });
},
$scope.pageChanged = function() {
   mainEvents(currentItem);
   var currentItem = ($scope.currentNumber - 1) * 10;
   $location.search('page', $scope.currentNumber);  //$scope.currentNumber is pageNumber
};

HTML

 <pagination total-items="totalItems" ng-model="currentNumber" ng-change="pageChanged()" items-per-page="itemsPerPage"></pagination>

角度ui-bootstrap分页不使用带有后退或前进按钮的URL

0 个答案: