Angularjs 角度分页反向跳转
我有这张桌子: 以下是代码和: HTMLAngularjs 角度分页反向跳转,angularjs,Angularjs,我有这张桌子: 以下是代码和: HTML <table class="table table-striped table-condensed table-hover"> <thead> <tr> <th class="id">Id <a ng-click="sort_by('id')"><i class="icon-sort"></i></
<table class="table table-striped table-condensed table-hover">
<thead>
<tr>
<th class="id">Id <a ng-click="sort_by('id')"><i class="icon-sort"></i></a></th>
<th class="name">Name <a ng-click="sort_by('name')"><i class="icon-sort"></i></a></th>
<th class="description">Description <a ng-click="sort_by('description')"><i class="icon-sort"></i></a></th>
<th class="field3">Field 3 <a ng-click="sort_by('field3')"><i class="icon-sort"></i></a></th>
<th class="field4">Field 4 <a ng-click="sort_by('field4')"><i class="icon-sort"></i></a></th>
<th class="field5">Field 5 <a ng-click="sort_by('field5')"><i class="icon-sort"></i></a></th>
</tr>
</thead>
<tfoot>
<td colspan="6">
<div class="pagination pull-right">
<ul>
<li ng-class="{disabled: currentPage == 0}">
<a href ng-click="prevPage()">« Prev</a>
</li>
<li ng-repeat="n in range(pagedItems.length, currentPage, currentPage + gap) "
ng-class="{active: n == currentPage}"
ng-click="setPage()">
<a href ng-bind="n + 1">1</a>
</li>
<li ng-class="{disabled: (currentPage) == pagedItems.length - 1}">
<a href ng-click="nextPage()">Next »</a>
</li>
</ul>
</div>
</td>
</tfoot>
<pre>pagedItems.length: {{pagedItems.length|json}}</pre>
<pre>currentPage: {{currentPage|json}}</pre>
<tbody>
<tr ng-repeat="item in pagedItems[currentPage] | orderBy:sortingOrder:reverse">
<td>{{item.id}}</td>
<td>{{item.name}}</td>
<td>{{item.description}}</td>
<td>{{item.field3}}</td>
<td>{{item.field4}}</td>
<td>{{item.field5}}</td>
</tr>
</tbody>
</table>
身份证件
pagedItems.length:{{pagedItems.length | json}
currentPage:{{currentPage | json}}
{{item.id}
{{item.name}
{{item.description}
{{item.field3}}
{{item.field4}
{{item.field5}}
JS
<li ng-repeat="n in range(pagedItems.length, currentPage - left_gap, currentPage + right_gap) ">
函数ctrlRead($scope,$filter){
//初始化
$scope.sortingOrder='name';
$scope.gap=5;
$scope.cached=0;
$scope.reverse=false;
$scope.filteredItems=[];
$scope.groupedItems=[];
$scope.itemsPerPage=5;
$scope.pagedItems=[];
$scope.currentPage=0;
$scope.items=[
{“id”:1,“name”:“name 1”,“description”:“description 1”,“field3”:“field3 1”,“field4”:“field4 1”,“field5”:“field5 1”},
{“id”:2,“name”:“name 2”,“description”:“description 1”,“field3”:“field3 2”,“field4”:“field4 2”,“field5”:“field5 2”},
//....
];
var searchMatch=函数(草堆、针){
如果(!针){
返回true;
}
返回haystack.toLowerCase().indexOf(needle.toLowerCase())!=-1;
};
//初始化筛选的项目
$scope.search=函数(){
$scope.filteredItems=$filter('filter')($scope.items,函数(item){
对于(项目中的var attr){
if(searchMatch(项[attr],$scope.query))
返回true;
}
返回false;
});
//注意分拣顺序
如果($scope.sortingOrder!=''){
$scope.filteredItems=$filter('orderBy')($scope.filteredItems,$scope.sortingOrder,$scope.reverse);
}
$scope.currentPage=0;
//现在按页面分组
$scope.groupToPages();
};
//计算页面位置
$scope.groupToPages=函数(){
$scope.pagedItems=[];
对于(变量i=0;i<$scope.filteredItems.length;i++){
如果(i%$scope.itemsPerPage==0){
$scope.pagedItems[Math.floor(i/$scope.itemsPerPage)]=[$scope.filteredItems[i]];
}否则{
$scope.pagedItems[Math.floor(i/$scope.itemsPerPage)].push($scope.filteredItems[i]);
}
}
};
$scope.range=函数(大小、开始、结束){
如果($scope.cached==start){
开始=开始-4;
console.log('start',start);
}
$scope.cached=start;
var-ret=[];
日志(大小、开始、结束);
if(size<2){return ret;}
如果(大小<结束){
结束=大小;
开始=大小-$scope.gap;
}
对于(变量i=开始;i<结束;i++){
如果(i 0){
$scope.currentPage--;
}
};
$scope.nextPage=函数(){
如果($scope.currentPage<$scope.pagedItems.length-1){
$scope.currentPage++;
}
};
$scope.setPage=函数(){
$scope.currentPage=this.n;
};
//函数描述了用于显示的数据处理
$scope.search();
//更改排序顺序
$scope.sort_by=函数(newSortingOrder){
如果($scope.sortingOrder==newSortingOrder)
$scope.reverse=!$scope.reverse;
$scope.sortingOrder=newSortingOrder;
};
};
ctrlRead.$inject=['$scope','$filter'];
从代码中可以看到,我们有13组5行
如果我在5上按“分页”,则5
按钮跳到第一位,最后一位是9
通过这种方式,我可以快速“浏览”所有数据
我的问题是,我不知道如何使它工作跳回a.e逆转
例如,如果我停留在13:
我按下9
我希望9
将跳转到分页的末尾,列表中的第一个元素将是5
如何做到这一点
谢谢,一般来说,您需要将当前页面索引和页面导航链接的索引分离。但是有很多方法可以做到这一点。我通过在您创建的范围中添加一个左右间距来实现这一点,这样可以更好地控制快速导航按钮的索引。因此,当您调用
设置页面
I只需添加一点检查:
if(这个.n
if (this.n <= $scope.currentPage) {
$scope.left_gap = $scope.gap-1;
$scope.right_gap = 1;
} else {
$scope.left_gap = 0;
$scope.right_gap = $scope.gap;
}
<li ng-repeat="n in range(pagedItems.length, currentPage - left_gap, currentPage + right_gap) ">