【问题标题】:Laravel - Jquery , Ajax - Inserting Nullable Data as an array in the databaseLaravel - Jquery,Ajax - 将 Nullable 数据作为数组插入数据库中
【发布时间】:2018-12-28 01:41:35
【问题描述】:

我正在尝试插入 2 个文件 imagevideo 以及作为数组的附加图像,我现在可以同时插入这两个文件,但是当我尝试将另一个图像添加为数组并保存时,输出也是插入在表单上选择的视频。如何将图像数组保存为独立图像并删除自动插入的视频

这是我的用户界面,请查看我的页面,我将其另存为图片并再次添加图片

这是我的数据库输出

我想要的输出是绿灯和黄灯应该是单一的,并且 没有插入视频,因为在页面上它们只是数组而不连接到视频

现在这是我的控制器代码

  $this->validate($request, [
        'paxsafety_image.*' => 'nullable|image|mimes:jpeg,jpg,png',
        'paxsafety_video.*' => 'nullable|mimes:mp4,mov,ogg | max:20000'
    ]);
    $paxSafety = [];
    $paxSafetyVideo = [];
    if ($request->has('paxsafety_image') && $request->has('paxsafety_video'))
    {   
        //Handle File Upload


        foreach ($request->file('paxsafety_image') as $key => $file)
        {
            // Get FileName
            $filenameWithExt = $file->getClientOriginalName();
            //Get just filename
            $filename = pathinfo( $filenameWithExt, PATHINFO_FILENAME);
            //Get just extension
            $extension = $file->getClientOriginalExtension();
            //Filename to Store
            $fileNameToStore = $filename.'_'.time().'.'.$extension;
            //Upload Image
            $path = $file->storeAs('public/paxsafety_folder',$fileNameToStore);
            array_push($paxSafety, $fileNameToStore);
        }



        foreach ($request->file('paxsafety_video') as $key => $file)
        {
            // Get FileName
            $filenameWithExt2 = $file->getClientOriginalName();
            //Get just filename
            $filename = pathinfo( $filenameWithExt2, PATHINFO_FILENAME);
            //Get just extension
            $extension2 = $file->getClientOriginalExtension();
            //Filename to Store
            $fileNameToStore2 = $filename.'_'.time().'.'.$extension2;
            //Upload Image
            $path = $file->storeAs('public/paxsafety_folder',$fileNameToStore2);
            array_push($paxSafetyVideo, $fileNameToStore2);
        }


        $fileNameToStore = serialize($paxSafety);
        $fileNameToStore2 = serialize($paxSafetyVideo);
    }
    else
    {
        $paxSafety[] = 'noimage.jpg';
        $paxSafetyVideo[] = 'noimage.jpg';
    }


    foreach ($paxSafety as $key => $value) {
        $paxSafetyContent = new PaxSafety;
        $paxSafetyContent->paxsafety_image = $value;
        foreach ($paxSafetyVideo as $key => $values) {
        $paxSafetyContent->paxsafety_video = $values;
        }
        $paxSafetyContent->save();
    }

这是我的查看代码

 {{ Form::file('paxsafety_image[]') }} &nbsp;&nbsp; <strong>  <span class="fa fa-camera"></span> Upload Image&nbsp;&nbsp; </strong>

 {{ Form::file('paxsafety_video[]') }} &nbsp;&nbsp; <strong>  <span class="fa fa-video-camera"></span> Upload Video&nbsp;&nbsp; </strong>

我的脚本代码

 <script>  
        $(document).ready(function(){  
             var i=1;  
             $('#add').click(function(){  
                  i++;  
                  $('#dynamic_field').append('<tr id="row'+i+'"><th>Upload new Image</th><td>{{ Form::file('paxsafety_image[]') }} <button type="button" name="remove" id="'+i+'" class="btn btn-danger btn_remove fa fa-minus-circle"></button></td></tr>');  
             });  
             $(document).on('click', '.btn_remove', function(){  
                  var button_id = $(this).attr("id");   
                  $('#row'+button_id+'').remove();  
             });  
             $('#submit').click(function(){  
            var form = $('#add_name')[0];
            var formData = new FormData(form);

              $.ajax({   
                   method:"POST",  
                   data:formData,  
                   success:function(data)  
                   {  
                        alert(data);  
                        $('#add_name')[0].reset();  
                   }  
              });  
         });

        });  
 </script>

另外,这是我的表架构

  $table->string('paxsafety_image')->nullable();
  $table->string('paxsafety_video')->nullable();

【问题讨论】:

  • 是您的upload new image and video 部分选择单个imagevideo 吗?

标签: javascript jquery mysql ajax laravel


【解决方案1】:

错误的部分在这里

foreach ($paxSafety as $key => $value) {
    $paxSafetyContent = new PaxSafety;
    $paxSafetyContent->paxsafety_image = $value;
  /*** 
   foreach ($paxSafetyVideo as $key => $values) {
    $paxSafetyContent->paxsafety_video = $values;
    }
  **/
    $paxSafetyContent->save();
}

您每次创建新的 PaxSafety 时都会插入视频 如果您确定只有一个视频附在第一张图片上,您可以这样做

foreach ($paxSafety as $key => $value) {
    $paxSafetyContent = new PaxSafety;
    $paxSafetyContent->paxsafety_image = $value;
   /** if it's the first image && the first video exists insert the first video **/
   if($key == 0 && isset($paxSafetyVideo[$key]) && $paxSafetyVideo[$key]){
    $paxSafetyContent->paxsafety_video = $paxSafetyVideo[$key];
    }

    $paxSafetyContent->save();
}

或者,如果您以后想附加带有多张图片的视频,也可以这样做

foreach ($paxSafety as $key => $value) {
    $paxSafetyContent = new PaxSafety;
    $paxSafetyContent->paxsafety_image = $value;
   /** if a video with the current image index exists insert the video value **/
   if(isset($paxSafetyVideo[$key]) && $paxSafetyVideo[$key]){
    $paxSafetyContent->paxsafety_video = $paxSafetyVideo[$key];
    }

    $paxSafetyContent->save();
}

【讨论】:

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