Question

我想在数据库中插入图像路径。在浏览器中，当任何用户选择图像路径时，该图像将存储在图像文件夹中，路径将存储在数据库中。我试过但if（$ type == ...）部分不起作用。它将转向else声明。

编辑：

if (isset($_POST['bSubmit'])) {
 var_dump($_FILES);
    $name = $_FILES['file']['name'];
     $type=var_dump($_FILES['file']['type']);



    if ($type == 'image/jpeg' || $type == 'image/png' || $type == 'image/gif') {
        if (file_exists(dirname($_SERVER['DOCUMENT_ROOT']) . '/htdocs/Amit_404_Store/img/' . $name)) {
            echo 'file is already present';
        } else {
            $uploadimage = move_uploaded_file($_FILES['file']['tmp_name'], dirname($_SERVER['DOCUMENT_ROOT']) . '/htdocs/Amit_404_Store/img/' . $name);
            echo "Stored in: " . "" . $name . "<br />";
            $destination = $name;
            echo "Upload in: " . "" . $destination . "<br />";

            //Database connection
            $servername = "Localhost";
            $username   = "root";
            $password   = "";
            $database   = "new_404_store2";

            //session value store and retrive data for index.html file
            $email = $_SESSION['user_name1'];

            $conn = mysqli_connect($servername, $username, $password, $database);

            if (mysqli_connect_errno()) {
                echo "Failed to connect to MySQL: " . mysqli_connect_error();
            }

            $Update_query = "update customer_information2 set Image_Path='$destination' where Email_Id='$email'";

            $result = mysqli_query($conn, $Update_query);



            if (!$result) {
                die('Could not enter data: ' . mysqli_error());
            }


            if ($uploadimage) {
                echo 'image uploaded and stored';
            } elseif (!$uploadimage) {
                echo 'image not uploaded';
            }

        }
    } else {
        echo 'Invalid file type';
    }
}

//mysqli_close($conn);
?>

Answer 1

您尚未编写enctype，它是文件传输所必需的

$user = Auth::user();
// services will be lazy loaded on first access of the attribute
$services = $user->services;
dd($user->toArray());

Answer 2

if ($type == 'image/jpeg' || $type == 'image/png' || $type == 'image/gif') {

应该是

if ($type == 'image/jpeg' || $type == 'image/jpg' || $type == 'image/png' || $type == 'image/gif') {

如果你正在添加图像文件但仍然出现错误

我想在PHP中使用MYSQL在数据库中插入图像路径

2 个答案: