Question

嗨，夥計們，我有一個問題，我上傳圖像上傳文件夾上傳工作正常，但他不提交價值到MySQL數據庫，我真的不知道失敗ist在這裏是整個代碼。 唯一值是來自用戶的ID，圖像名稱的字段是company_logo。 我的儀表板代碼： 的形式爲：

<form id="imageform" method="post" enctype="multipart/form-data" action='ajaximage.php'> 
          <input type="file" name="photoimg" id="photoimg" /> 
          </form> 

jQuery代碼

<script type="text/javascript" > 
$(document).ready(function() { 

      $('#photoimg').on('change', function()   { 
         $("#preview").html(''); 
       $("#preview").html('<div class="spinner"></div>'); 
      $("#imageform").ajaxForm({ 
         target: '#preview' 
     }).submit(); 

      }); 
     }); 
</script> 

終於ajaximage.php

<?php 
session_start(); 
ob_start(); 
$valid_user_id = trim($_SESSION["VALID_USER_ID"]); 

if(isset($_SESSION["VALID_USER_ID"]) && !empty($valid_user_id)) 
{ 

include "database_connection.php"; //Include the database connection script 

//Check the logged in user information from the database 
$check_user_details = mysql_query("select * from `signup_and_login_table` where `email` = '".mysql_real_escape_string($_SESSION["VALID_USER_ID"])."'"); 

//Get the logged in user info from the database 
$get_user_details = mysql_fetch_array($check_user_details); 

//Pass all the logged in user info to variables to easily display them when needed 
$user_id = strip_tags($get_user_details['id']); 
$firstname = strip_tags($get_user_details['firstname']); 
$lastname = strip_tags($get_user_details['lastname']); 
$company = strip_tags($get_user_details['company']); 
$company_logo = strip_tags($get_user_details['company_logo']); 
$email = strip_tags($get_user_details['email']); 
$passwd = strip_tags($get_user_details['password']); 

// User Id for Image Upload 
$session_id = strip_tags($get_user_details['id']); 


$path = "uploads/"; 

$valid_formats = array("jpg", "png", "gif", "bmp"); 
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST") 
    { 
     $name = $_FILES['photoimg']['name']; 
     $size = $_FILES['photoimg']['size']; 

     if(strlen($name)) 
      { 
       list($txt, $ext) = explode(".", $name); 
       if(in_array($ext,$valid_formats)) 
       { 
       if($size<(1024*1024)) 
        { 
         $actual_image_name = time().substr(str_replace(" ", "_", $txt), 5).".".$ext; 
         $tmp = $_FILES['photoimg']['tmp_name']; 
         if(move_uploaded_file($tmp, $path.$actual_image_name)) 
          { 
          mysqli_query($db,"UPDATE signup_and_login_table SET company_logo='$actual_image_name' WHERE id='$session_id'"); 

           echo "<img src='uploads/".$actual_image_name."' class='preview'>"; 
          } 
         else 
          echo "failed"; 
        } 
        else 
        echo "Image file size max 1 MB";      
        } 
        else 
        echo "Invalid file format.."; 
      } 

     else 
      echo "Please select image..!"; 

     exit; 
    } 

} 
else 
{ 
//Send every user who tries to access this page directly without valid session to the login page. 
//The login page is the door that every user needs to pass to this page 
header("location: login.html"); 
} 
?> 

Answer 1

使用下面的SQL查詢： INSERT INTO signup_and_login_table (company_logo, id) VALUES ('$actual_image_name', '$session_id')

您我們做了一個UPDATE查詢，它只更新已存在的行。

親切的問候！

Answer 2

嘗試此查詢

mysqli_query($db,"UPDATE signup_and_login_table SET company_logo='$actual_image_name' WHERE id=".$session_id);