在角度ui-grid中选择行时,仅获取可见列

时间:2017-04-04 13:04:29

标签: javascript angularjs angular-ui-grid ui-grid

我想在angular-ui网格中选择一行并将该行复制到剪贴板。

这是我的代码:

  $scope.copySelection = function() {
    $scope.retainSelection = $scope.gridApi.selection.getSelectedRows();
    alert(JSON.stringify($scope.retainSelection));
    var input = document.createElement("input");
    input.type = "text";
    document.getElementsByTagName('body')[0].appendChild(input);
    input.value = JSON.stringify($scope.retainSelection);
    input.select();
    document.execCommand("copy");
    input.hidden = true;
    $scope.gridApi.selection.clearSelectedRows();
  };

Plunker:http://plnkr.co/edit/dcj7DUWHyA3u1bouxRhI?p=preview

但是,我只想复制可见列,但我得到了JSON中的所有列。我不想要隐藏的列。我怎么做?请帮忙。

1 个答案:

答案 0 :(得分:2)

您可以在所选列/可见列的基础上调整列。你可以有这样的代码 -

 $scope.copySelection = function() {

    $scope.retainSelection =angular.copy($scope.gridApi.selection.getSelectedRows());

    angular.forEach($scope.retainSelection,function(value,key){
       var columndef=angular.copy( $scope.gridOptions.columnDefs);
      for (var property in value) {
       if (!(value.hasOwnProperty(property) && columndef.filter(function(a){return a.name.split('.')[0]===property}).length>0 )) {
        delete value[property];
      }
    }

    });
    alert(JSON.stringify($scope.retainSelection));
    var input = document.createElement("input");
    input.type = "text";
    document.getElementsByTagName('body')[0].appendChild(input);
    input.value = JSON.stringify($scope.retainSelection);
    input.select();
    document.execCommand("copy");
    input.hidden = true;
    $scope.gridApi.selection.clearSelectedRows();
  };

查找更新的Plunker Here

希望它能解决你的问题!