将图像上传到服务器并将url保存到服务器

Question

我正在构建一个图像上传器，我想通过该图像上传器将图像上传到服务器，并用新的图像替换数据库中的旧图像URL。上传部分工作正常，但我无法在数据库中获取imageURL。有人可以看看我的代码，并告诉我我在哪里做错了吗？

<?php 
$target_dir = "media/images/"; 
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]); 
$uploadOK = 1; 
$imageFileType = pathinfo($target_file, PATHINFO_EXTENSION); 
// Check if image file is a actual image or fake image 
if(isset($_POST["uploadImage"])) { 
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]); 
     if ($check != false) { 
     echo "File is an image - " .$check["mime"]. "."; 
     $uploadOK = 1; 
     } else { 
     echo "File not an image"; 
     $uploadOK = 0; 
    } 
} 

// check if file exists 
if (file_exists($target_file)) { 
    echo "sorry File exists"; 
     $uploadOK = 0; 
} 

    // check fle size 
    if ($_FILES["fileToUpload"]["size"] > 5000000) { 
    echo "Sorry your file is too large."; 
    $uploadOK = 0; 
} 

// Allow certain file formats 
if ($imageFileType != "jpg" && $imageFileType != "png" &&  $imageFileType != "jpeg" && $imageFileType != "gif") { 
echo "sorry, only JPG, JPEG, PNG and GIF files allowed."; 
$uploadOK == 0; 
} 

// check if $uploadOK is set 0 by an error 
if ($uploadOK == 0) { 
    echo "Sorry your file was not uploaded."; 
    // if everything is ok try to upload file 
    } else { 
    if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) { 
    echo "The file ". basename($_FILES["fileToUpload"]["name"]). " has  been uploaded."; 
    include 'connect.php'; 

    $stmt = $conn->prepare("INSERT INTO image(name,imageURL, image_cat_id, AlbumID) VALUES (home1, ?, 8, 0)"); 

    $stmt->bind_param('s', $target_file); 
    $stmt->execute(); 
    $stmt->close(); 
} else { 
    echo "Sorry, there was an error uploading your file."; 
} 

} 
?> 

结果在成功上传：

File is an image - image/png.The file arrowdown_51.png has been uploaded. Fatal error: Call to a member function bind_param() on boolean in C:\xampp\htdocs\WebDevelopment\Mosta Dynamic\CMS\upload.php on line 47

Answer 1

这是可能的，该功能准备（）返回一个布尔值...我猜，您所做的SQL查询是错误的。具有以下尝试：

INSERT INTO image(name,imageURL, image_cat_id, AlbumID) 
VALUES ('home1', ?, '8', '0') 

编辑：如果上面的查询也没有持续的记录：还要检查，如果你的列（并表）的名称是正确的。

请原谅我的英语不好，我不是一个母语... :)