Question

我正在尝试过滤并仅显示未删除的用户（有效）其中let appDelegate = UIApplication.shared.delegate as! AppDelegate appDelegate.window?.rootViewController = newController是表，users是列（时间戳）。

deleted_at

我在这里做什么错了？

Answer 1

尝试这样。我希望它能为您提供帮助。

<tr ng-repeat="user in users | filter: {!user.deleted_at}" ng-class="{'success': userData.id == user.id }" style="background-color:{{'userData.color'}}" ng-if="selectedRole == '' || (user.group_list | lookup:GROUPS | lists:'name' | join).indexOf(selectedRole) != -1">

Answer 2

只需使用filter : {deleted_at : null}

function myctrl($scope){
  $scope.users = [
    {"name":"Hardik Shah", "deleted_at" : null},
    {"name" : "I am deleted", "deleted_at": "2343434"},
    {"name": "I am not deleted", "deleted_at": null}
  ]
}

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<table ng-app = "" ng-controller="myctrl">
<tr ng-repeat="user in users | filter : {deleted_at : null}">
  <td>
    {{user.name}}
  </td>
</tr>
</table>

Answer 3

您可以创建一个函数来过滤数组并将其传递给ng-repeat

function ctrl($scope){
  $scope.users = [
    {id: 1, deleted_at: null},
    {id: 2, deleted_at: "12345"},
    {id: 3, deleted_at: null}
  ]
  $scope.userFilter = user => user.deleted_at
}

<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<table ng-app ng-controller="ctrl">
  <tr ng-repeat="user in users | filter: userFilter">
    <td>
      {{user.id}}
    </td>
  </tr>
</table>

AngularJS：仅过滤未删除的用户

3 个答案: