如何在輸入文件中發送動態值？

Question

我創建了一個編輯表單，我正在編輯我的房間號碼和房間描述以及房間圖像等。基於類別ID。

第一個編輯表單將從數據庫中獲取數據。 示例代碼：

if (isset($_GET['id'])) 
       { 
        $current_id = $_GET['id']; 
        $query=mysql_query("select * from room where room_id='$current_id'"); 
         $i = 0; 
         while($row=mysql_fetch_array($query)) 
         { 
          $i++; 
          $room_id=$row['room_id']; 
          $cat_id=$row['cat_id']; 
          $room_number=$row['room_no']; 
          $room_desc=$row['room_desc']; 
          $room_image=$row['room_image']; 

        ?> 
        <table class="table table-bordered table-hover table-striped"> 
          <thead> 
           <tr> 
            <th>Room Id</th> 
            <th>Category Type</th> 
            <th>Room Number</th> 
            <th>Room Description</th> 
            <th>Room Image</th> 

           </tr> 
          </thead> 
          <tbody> 
          <tr class="active"> 
            <td><input type="text" value="<?php echo $room_id; ?>" readonly="readonly" name="room_id"></td> 
            <td> 

            <?php   
               // Slecting cat_id for displaying cat_type on Edit form 
            $query_room=mysql_query("SELECT * FROM `category` WHERE cat_id in (". $cat_id .")"); 

            while($row=mysql_fetch_array($query_room)) 
            { 
             $category_name=$row['cat_type']; 
             $category_id=$row['cat_id']; 
             echo '<input type="text" value='.$category_name.' readonly="readonly" name="cat_name">'; 

             echo '<input type="hidden" value='.$category_id.' readonly="readonly" name="cat_id">'; // hidden value for cat_id to fetch data from category 
             $_SESSION['imagesession'] = "data:image/png;base64, base64_encode($room_image)"; 
            } 
            ?> 
            </td> 
            <td><input type="text" value="<?php echo $room_number ;?>" name="room_number" required></td> 
            <td><input type="text" value="<?php echo $room_desc ;?>" name="room_desc" required></td> 
            <td><input type="image" src="data:image/png;base64,<?php echo base64_encode($room_image);?>" name="myimage" alt="No Photo" title="Click to View in Detail" height="50px" width="50px"/> 

            <label>Upload Image</label> 
             <input type="file" id="uploadImage" value="<?php echo base64_encode($room_image);?>" name="image" onChange="PreviewImage();" /> 
             <img id="uploadPreview" style="width: 200px; height: 300px; border-style:none" /> 

            </td> 
            </tr> 
           </tbody> 
     </table> 
         </div> 
         </div> 
         <button type="submit" class="btn btn-success" name="update_rooms">Update Rooms</button> 
         <button type="reset" class="btn btn-danger">Reset</button> 
         </form> 

獲取數據，並在數據庫中更新

  if(isset($_POST['update_rooms'])) 
      { 
      $current_id = $_POST['room_id']; 
      $room_desc = $_POST['room_desc']; 
      $room_number = $_POST['room_number'];    
      $cat_id = $_POST['cat_id']; 


      // $imageName = mysql_real_escape_string($_FILES["image"]["name"]); 
      $imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"])); 
      $imageType = mysql_real_escape_string($_FILES["image"]["type"]); 
      if(empty($imageData)) 
      { 
       if($imageData === $existing_category_image) 
       { 
       echo '<span style="color:#E02F2F;text-align:center;font-size:30px;padding-left:190px;">Reselect the Image</span>';     
       return; 
       } 
      } 

      $sql = "UPDATE room SET cat_id='$cat_id', room_no='$room_number', room_desc='$room_desc', room_image='$imageData' WHERE room_id='$current_id'"; 
      //echo "Res".$sql; 
      $result=mysql_query($sql); 
      if($result) 
      { 
      echo '<span style="color:#E02F2F;text-align:center;font-size:30px;padding-left:190px;">Record Updated Successfully!</span>'; 
      //header("Location: success.php"); 
      } else 
      { 
      echo '<span style="color:#E02F2F;text-align:center;font-size:30px;padding-left:250px;">Error Found!.mysql_error()</span>'; 
      } 
} 

問題：

1.當編輯的所有字段更新成功，但來到圖像它不工作我的意思是，如果用戶正在添加新圖像，然後在數據庫中進行更新，但如果用戶正在編輯現有圖像，則空值將存儲在數據庫中。

如何解決這個問題。如何在輸入類型文件中發送動態值。任何幫助？

Answer 1

你也可以使用這個

取這個$existing_category_image從數據庫不需要從形式張貼。

if($imageData == '' || $imageData == null){ 
$sql = "UPDATE room SET cat_id='$cat_id', room_no='$room_number', room_desc='$room_desc' WHERE room_id='$current_id'"; 
}else{ 
$sql = "UPDATE room SET cat_id='$cat_id', room_no='$room_number', room_desc='$room_desc', room_image='$imageData' WHERE room_id='$current_id'"; 
} 

OR

這樣寫：

if($imageData == '' || $imageData == null){ 
      $imageData = $existing_category_image; 
    } 

前

$sql = "UPDATE room SET cat_id='$cat_id', room_no='$room_number', room_desc='$room_desc', room_image='$imageData' WHERE room_id='$current_id'";