javascript - Laravel-如果我将变量传递给 Controller ​​并保存在数据库中，如何在 javascript 中使用 switch case

Question

我在 laravel 工作，我发送了很多数据，包括下拉列表，它们有值，0,1,2，这意味着顶部，右侧，左侧问题是当我仅在 Controller 中使用更新功能时，我得到 0 或 1 或 2取决于我输入的内容，必须刷新才能看到文本，因为我在单击保存后使用 json 和 ajax。

HTML

<select class="form-control" id="position_edit" name="position_edit">
                <option value="0">TOP</option>
                <option value="1">RIGHT</option>
                 <option value="2">LEFT</option>
                <option value="3">BOTTOM</option>
               </select>

控制者

public function updatebanner(Request $request)
{
    if ($request->hasFile('image')) {
    $imagename = time().'.'.$request->file('image')->getClientOriginalExtension();
    $path =$request->file('image')->storeAs('/images', $imagename);

    $post = Advertisement::findOrFail($request->id);
    $post->adver_title =  $request->title_edit; 
    $post->adver_url =  $request->url_edit; 
    $post->adver_position =  $request->position_edit; 
    $post->adver_photo = $imagename; 
    $post->save();
}

ajax 代码

$('.item' + data.id).replaceWith("
<tr class='item" + data.id + "'>
   <td  class='col1'>" + data.id + "</td>
   <td><img src='images/" + data.adver_photo + "'style='width: 50%;margin-left: 35px;border-radius: 3px;border: 1px solid #1a2732;' /> </td>
   <td>" + data.adver_title + "</td>
   <td>" +data.adver_position+ "</td>
   </td>
   <td class='text-center'><input type='checkbox' class='edit_published' data-id='" + data.id + " '></td>
   <td style='text-align:center;'>Right Now!</td>
   <td><button class='show-modal btn btn-success' data-id='" + data.id + "' data-titlefield='" + data.adver_title + "' data-photofield='images/" + data.adver_photo + "'><span class='fa fa-eye'></span> Show</button> <button class='edit-modal btn btn-info' data-id='" + data.id + "' data-titlefield='" + data.adver_title + "' data-photofield='images/" + data.adver_photo + "'><span class='fa fa-pencil-square'></span> Edit</button> <button class='delete-modal btn btn-danger' data-id='" + data.id + "' data-titlefield='" + data.adver_title + "' data-photofield='images/" + data.adver_photo + "'><span class='fa fa-trash-o'></span> Delete</button></td>
</tr>
");

我尝试使用下一个代码，但如果 adver_position 值保存在数据库中，问题就不起作用，它将返回空白，没有任何数字或文本，必须刷新页面才能看到文本

   var content = "<tr class='item" + data.id + "'> "
            +  "<td  class='col1'>" + data.id + "</td> "
            +  "<td><img src='images/" + data.adver_photo + "'style='width: 50%;margin-left: 35px;border-radius: 3px;border: 1px solid #1a2732;' /> </td>" 
            +  "<td>" + data.adver_title + "</td>"
            +  "<td> ";

    switch(data.adver_position) {
case 0:
    content = content + "TOP";
    break;
case 1:
    content = content + "RIGHT";
    break;
case 2:
    content = content + "LEFT";
    break;
case 3:
    content = content + "BOTTOM";
    break;
}

content = content + "</td> </tr>"


 $('.item' + data.id).replaceWith(content);

Answer 1

根据我对你的问题的理解，我认为当你从你的 Controller 保存到数据库时，你不会向前端返回响应，这可以用来更新你的内容。

为什么不做这样的事情呢？

public function updatebanner(Request $request)
  {
 if ($request->hasFile('image')) {
 $imagename = time().'.'.$request->file('image')->getClientOriginalExtension();
$path =$request->file('image')->storeAs('/images', $imagename);

$post = Advertisement::findOrFail($request->id);
$post->adver_title =  $request->title_edit; 
$post->adver_url =  $request->url_edit; 
$post->adver_position =  $request->position_edit; 
 $post->adver_photo = $imagename; 
 $post->save();
return response(['item'=>$post]); /// return response

}