Question

我有一个用于上传图片的php文件，重命名它，因此它有一个独特的值，可以获取其他用法，上传部分工作，但当我采取变量（$ fileNewName）并尝试将其插入到数据库，它没有用。

session_start();
include_once('dbh-inc.php');
$file = $_FILES['file'];

$fileName = $_FILES['file']['name'];
$fileTmpName = $_FILES['file']['tmp_name'];
$fileSize = $_FILES['file']['size'];
$fileError = $_FILES['file']['error'];
$fileType = $_FILES['file']['type'];

$fileExt = explode('.', $fileName);
$fileActualExt = strtolower(end($fileExt));


$allowed = array('jpg', 'jpeg', 'png', 'pdf');

if (in_array($fileActualExt, $allowed)) {
    if ($fileError === 0) {
        if ($fileSize < 1000000) {
            $fileNameNew = uniqid('',true).".". $fileActualExt;
            $fileDestination = '../uploads/' .$fileNameNew;

            move_uploaded_file($fileTmpName,$fileDestination);
            $sql = "UPDATE stenden_users SET userImagePath = ".$fileNameNew." WHERE userID =" .$_SESSION['u_id'] ; 
            if ($stmt = mysqli_prepare($conn,$sql)) {
                $result = mysqli_stmt_execute($stmt);
                if($result == FALSE){
                    echo "failed";
                } else {
                    echo "success!!!";
                }


            }


            echo "Your image was successfully uploaded! ";
        } else {
            echo "The size of file is too big";
        }
    } else {
        echo "There was an error uploading your file";
    }
} else {
    echo "You cannot upload files of this type";
}

Answer 1

您的SQL语法不正确，您可以将SQL查询更改为：

$sql = "UPDATE stenden_users SET userImagePath = '".$fileNameNew."' WHERE userID =" .$_SESSION['u_id'] ;

无法在数据库中存储文件名

1 个答案: