Question

我正在编写一个脚本，只有在python文件的输出显示dup_yes为1后才会打开模态

我的模态如下：

<button type="button" class="btn btn-primary" data-toggle="modal" data-target="#dupModal" data-keyboard="false" data-backdrop="static" ng-click="save_me()" >Save Me</button>

<div class="modal fade" id="dupModal" role="dialog">
    <div class="modal-dialog">
        <!-- Modal content-->
        <div class="modal-content">
            <div class="modal-header">
                <h3 class="modal-alert">Same name already exists</h3>
                <h6 class="modal-ask">Save both entries or do you want to cancel?</h6>
            </div>
            <div class="modal-footer">
                <button type="button" class="btn btn-default" data-dismiss="modal" aria-hidden="true" >Keep Both</button>
                <button type="button" class="btn btn-default" data-dismiss="modal" aria-hidden="true" >Close</button>
            </div>
        </div>
    </div>
</div>

我的脚本如下：

$scope.save_me = function () {
    http.get("path to python file").then(function(response){
        $scope.check=response.data;
        if ($scope.check['login_status'] == 1 ){
           console.log("Mission is a Success\n");
           $scope.dup_yes=1;
        }
    })
}

当我点击按钮时，它会打开模态并同时运行python文件。

相反，我希望它只在我的python脚本给出值1时打开模态。

关于如何做到这一点的任何想法？

Answer 1

您可以从控制器访问模态并将其打开，如下所示：

$scope.save_me = function () {
   http.get("path to python file").then(function(response){
   $scope.check=response.data;
       if ($scope.check['login_status'] == 1 ){
          console.log("Mission is a Success\n");
          $scope.dup_yes=1;

          var modal = angular.element('#dupModal');
          //modal.modal('hide');
          modal.modal('show');

       }
   })
  }

Mayby更好的解决方案是在separete控制器中处理模态操作并将其注入当前控制器以将责任分开

仅当数据为真时才打开模态

1 个答案: