Question

我正在尝试将一些数据上传到我的网站中。我尝试上传的图像已上传到我的图像文件夹中，但数据库没有输入数据。它在mysqli_stmt_bind_param中显示错误。我找不到错误。有人可以帮我吗？

if (isset($_POST['submit1'])) {
    $cakename = $_POST['cakeName'];

    $Beschreibung = $_POST['beschreibung'];
    $Preis = $_POST['preis'];

    $file = $_FILES['image'];

    $fileName = $file["name"];
    $filetype = $file["type"];
    $fileTempName = $file["tmp_name"];
    $fileError = $file["error"];
    $fileSize = $file["size"];

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

    $allowed = array("jpg", "jpeg", "png");

    if (in_array($fileActualExt, $allowed)) {
        if ($fileError === 0) {
            if ($fileSize < 2000000) {
                $imageFullName = $cakename.".". uniqid("", true).".". $fileActualExt;
                $fileDestination = "../Bake_my_cake1/image/".$imageFullName;

                include_once 'lib/db_connector.php';
                $db = dbconnect();

                if (empty($Beschreibung) || empty($Preis)) {
                    header("Location: ../Bake_my_cake1/Profile_baker1.php?upload=empty");
                    exit();
                } else {
                    $sql = "SELECT * FROM Cake;";
                    $stmt = mysqli_stmt_init($db);
                    if (!mysqli_stmt_prepare($stmt, $sql)) {
                        echo "SQL statement failed 1";
                    } else {
                        mysqli_stmt_execute($stmt);
                        $result = mysqli_stmt_get_result($stmt);
                        $rowCount = mysqli_num_rows($result);
                        $setImageOrder = $rowCount + 1;

                        $sql = "INSERT INTO Cake (CakeName, Beschreibung, Preis, orderGallery) VALUES(?, ?, ?, ?);";

                        if (!mysqli_stmt_prepare($stmt, $sql)) {
                            echo "SQL statement failed 2";
                        } else {
                            mysqli_stmt_bind_param($stmt, "ssss", $imageFullName, $Beschreibung, $Preis, $setImageOrder);
                            mysqli_stmt_execute($stmt);
                            echo "error 1";
                            move_uploaded_file($fileTempName, $fileDestination);

                            header("Location ../Bake_my_cake1/Profile_baker1.php?upload=sucess");
                        }
                    }
                }
                dbclose($db);
            } else {
                echo"File Size is too Big";
                exit();
            }
        } else {
            echo"You have an error";
            exit();
        }
    } else {
        echo"You need to upload a proper file type";
        exit();
    }
}

?>

else{
    mysqli_stmt_bind_param($stmt,"ssss", $imageFullName, $Beschreibung, $Preis, $setImageOrder);
    mysqli_stmt_execute($stmt);
    echo "error 1";
    move_uploaded_file($fileTempName,$fileDestination);

    header("Location ../Bake_my_cake1/Profile_baker1.php?upload=sucess");

它显示错误1。

Answer 1

根据您的变量名，我假设您的函数dbconnect()返回数据库对象。并且您正在$db中传递那个mysqli_stmt_init而不是数据库连接对象。

例如。

如果您要返回数据库连接对象并将其存储在$con中，那么您的代码将类似于$stmt = mysqli_stmt_init($con)。

使用此代码，您不会看到"SQL statement failed 1"消息，并且您的代码将按预期执行。

插入的数据不会进入我的数据库

1 个答案: