Laravel-如果我將變量傳遞給控制器​​並保存在數據庫中，如何在 javascript 中使用 switch case

Question

我在 laravel 工作，我發送了許多數據，包括具有值的下拉列表，0,1,2，這意味着頂部，右側，左側問題是當我只在控制器中使用更新功能時，我得到 0 或 1 或 2 取決於什么我輸入並必須刷新才能看到文本，因為我在單擊保存后使用 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>

控制器

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中的代碼

$('.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>
");

我嘗試使用下一個代碼，但如果 adver_position 值保存在數據庫中，則問題不起作用，它將返回空白，沒有任何數字或文本，必須刷新頁面才能看到文本

   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

根據我對您的問題的理解，我認為當您從控制器保存到數據庫時，您不會向前端返回響應，該響應可用於更新您的內容。

為什么不做這樣的事情？

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

}

Answer 2

你可以試試：

content += $('#position_edit option:checked').text();

而不是switch 。 這將簡單地分配所選位置的文本。

編輯：替換所有這些：

    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;
}

使用提供的解決方案。