將圖像上傳到服務器並將url保存到服務器

Question

我正在構建一個圖像上傳器，我想通過該圖像上傳器將圖像上傳到服務器，並用新的圖像替換數據庫中的舊圖像URL。上傳部分工作正常，但我無法在數據庫中獲取imageURL。有人可以看看我的代碼，並告訴我我在哪裏做錯了嗎？

<?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."; 
} 

} 
?> 

結果在成功上傳：

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

這是可能的，該功能準備（）返回一個布爾值...我猜，您所做的SQL查詢是錯誤的。具有以下嘗試：

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

編輯：如果上面的查詢也沒有持續的記錄：還要檢查，如果你的列（並表）的名稱是正確的。

請原諒我的英語不好，我不是一個母語... :)