我得到“ else”，而不是“ if”

Question

代码操作

<?php 
    $link = mysqli_connect("localhost", "root", "usbw", "homelessons");
    mysqli_set_charset($link, "utf8");
?>
<?php 
    if(isset($_POST["add"])){
        $name = $_POST["name"];
        $price = $_POST["price"];
        $img = "";
        $date = $_POST["date"];

        if(isset($_FILES["image"])){
            $img_folder = "";
            $image_name = $_FILES["image"]["name"];
            $img = $img_folder . $image_name;
            $file_path = "/img" . $img;

            move_uploaded_file($_FILES["image"]["tmp_name"], $file_path);
        }
        $sql_add = "
            INSERT INTO `lesson_first`
                (name, price, img, date)
            VALUES
                ('$name', '$price', '$img', '$date')
        ";
        $func = mysqli_query($link, $sql_add);

        if($func == True){
            header("Location: show.php");
        }
        else{
            echo "Something went wrong";
            mysqli_error($link);
        }
    }
?>

代码形式

    <form action="add.php" method="post" enctype="multipart/form-data">
        <label for="name">name</label><br>
        <input type="text" name="name"><br><br>

        <label for="price">price</label><br>
        <input type="number" name="price" rows="10" cols="46"><br><br>

        <input type="file" name="image"><br><br>

        <input type="date" name="date"><br><br>

        <input type="submit" name="add">
    </form>

我有将从表单接收到的信息添加到sql数据库的代码，并且看来我写的都正确，因为我没有任何错误，但是我得到了另外的提示：“出了点问题”，所以我不知道在哪里我的错误

Answer 1

尝试一下：

if(! $func){
   echo "Something went wrong";
   mysqli_error($link);
}else{
   header("Location: show.php");
}