用php创建的网站未保存在db中[关闭]

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关闭。 这个问题需要调试细节。 它目前不接受答案。

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3 天前关闭。

改进这个问题

<meta charset="utf-8">
<?php
session_start();

$mn = $_POST['moutain'];
$ny = $_POST['naeyong'];

$mn = htmlspecialchars($mn, ENT_QUOTES);
$ny = htmlspecialchars($ny, ENT_QUOTES);

$day = date("Y-m-d (H:i)");

$upload = "./data/";

$upimage_name = $_FILES["upimage"]["name"];
$upimage_tmp_name = $_FILES["upimage"]["tmp_name"];
$upimage_type = $_FILES["upimage"]["type"];
$upimage_size = $_FILES["upimage"]["size"];
$upimage_error = $_FILES["upimage"]["error"];

if($upimage_name && !$upimage_error){
    $image = explode(".", $upimage_name);
    $image_name = $image[0];
    $image_ext = $image[1];

    $new_image_name = date("Y_m_d_h_i_s");
    $new_image_name = $new_image_name;
    $copied_image_name = $new_image_name.".".$image_ext;
    $uploaded_image = $upload.$copied_image_name;

    if($upimage_size > 1000000){
        echo ("<script>
                alert('The capacity has been exceeded. Please check again.');
                history.go(-1)
            </script>
            ");
            exit;
    }
}

else{
    $upimage_name = "";
    $upimage_type = "";
    $copied_image_name = "";
}
$con = mysqli_connect("localhost", "secure", "secure", "secure") or die("not connect!");
$sql = "insert into test1 (date, mountain, ny, image_name, image_type, image_copied) ";
$sql = "value('$day', '$mn', '$ny', '$upimage_name', '$upimage_type', '$copied_image_name')";
mysqli_query($con, $sql);

mysqli_close($con);

echo "<meta http-equiv='refresh' content='0;url=main.php'>";
?>

在某座山上，什么时候，发生了什么，我正在做一个网站来查看和写下写这篇文章的文章。 我试图连接到数据库并运行它，但我无法保存到数据库。 但是，没有出现其他错误。 这段代码有错吗？