Question

如何从网格中过滤这些行时取消选择行？在angular-ui-grid中有没有可以完成这项工作的旗帜？类似的东西：

function A = pushmatrixup(A)
    % Loop over each column of matrix A
    for col = 1:size(A,2)       
        % Rearrange the values in column 'col', with the logic:
        % [Values in column 'col' which don't equal 0;
        %  Values in column 'col' which do equal 0]
        A(:,col) = [A(A(:,col) ~= 0, col); A(A(:,col) == 0, col)];
    end
    % All columns now have their 0s 'sunk' to the bottom of the column.
end

Answer 1

如果只有gridOptions.removeSelectedAfterFilterd = true。

这是一个有效的替代方案：

＆＃13;

var app = angular.module('app', ['ui.grid', 'ui.grid.selection']);
app.controller('MainCtrl', ['$scope', '$timeout',
  function($scope, $timeout) {
    $scope.gridOptions = {
      enableFiltering: true,
      enableSelectAll: true,
      enableRowSelection: true,
      enableRowHeaderSelection: false,
      enableFullRowSelection: true,
      showGridFooter: true,
      onRegisterApi: function(gridApi) {
        $scope.gridApi = gridApi;
        $scope.gridApi.core.on.filterChanged($scope, function() {
          $timeout(function() {
            angular.forEach($scope.gridApi.selection.getSelectedGridRows(), function(row) {
              if (!row.visible) $scope.gridApi.selection.unSelectRow(row.entity);
            });
          }, 10);
        });
      },
      columnDefs: [{name: 'FirstName'}, {name: 'LastName'}, {name: 'Job'}],
      data: [{"FirstName": "nir", "LastName": "OP", "Job": "Stack Overflow User"},
             {"FirstName": "Tim", "LastName": "Harker", "Job": "Stack Overflow User"},
             {"FirstName": "nir", "LastName": "OP", "Job": "Stack Overflow User"}]
    };
  }
]);

＆＃13;

div[ui-grid] {
  height: 190px;
}

＆＃13;

<script src="//cdnjs.cloudflare.com/ajax/libs/angular.js/1.6.1/angular.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/angular-ui-grid/4.0.2/ui-grid.min.js"></script>
<link rel="stylesheet" href="//cdnjs.cloudflare.com/ajax/libs/angular-ui-grid/4.0.2/ui-grid.min.css" />
<div ng-app="app" ng-controller="MainCtrl">
  <div ui-grid="gridOptions" ui-grid-selection class="grid"></div>
</div>

＆＃13;

如果您有任何其他问题，请告诉我，我们乐意为您提供帮助。

Answer 2

就像Tim Harker的回答一样，无需添加“$ timeout”

var app = angular.module('app', ['ui.grid', 'ui.grid.selection']);
app.controller('MainCtrl', ['$scope', '$timeout',
  function($scope, $timeout) {
    $scope.gridOptions = {
      enableFiltering: true,
      enableSelectAll: true,
      enableRowSelection: true,
      enableRowHeaderSelection: false,
      enableFullRowSelection: true,
      showGridFooter: true,
      onRegisterApi: function(gridApi) {
        $scope.gridApi = gridApi;
        $scope.gridApi.core.on.rowsVisibleChanged($scope, function() {
            angular.forEach($scope.gridApi.selection.getSelectedGridRows(), function(row) {
              if (!row.visible) {
                  $scope.gridApi.selection.unSelectRow(row.entity)
                  };
            });

        });
      },
      columnDefs: [{name: 'FirstName'}, {name: 'LastName'}, {name: 'Job'}],
      data: [{"FirstName": "nir", "LastName": "OP", "Job": "Stack Overflow User"},
             {"FirstName": "Tim", "LastName": "Harker", "Job": "Stack Overflow User"},
             {"FirstName": "nir", "LastName": "OP", "Job": "Stack Overflow User"}]
    };
  }
]);

div[ui-grid] {
  height: 190px;
}

<script src="//cdnjs.cloudflare.com/ajax/libs/angular.js/1.6.1/angular.min.js"></script>
<script src="//cdnjs.cloudflare.com/ajax/libs/angular-ui-grid/4.0.2/ui-grid.min.js"></script>
<link rel="stylesheet" href="//cdnjs.cloudflare.com/ajax/libs/angular-ui-grid/4.0.2/ui-grid.min.css" />
<div ng-app="app" ng-controller="MainCtrl">
  <div ui-grid="gridOptions" ui-grid-selection class="grid"></div>
</div>

UI-GRID - 在行筛选后删除选定的行

2 个答案: