所以,我一直想知道这个剧本,但仍然无法做到正确。由于某种原因,它不会保存到我的数据库。任何想法为什么它不起作用?非常感谢任何帮助。谢谢!这是我的剧本。
<?php
include_once ("database.php"); ?>
<?php
if (isset($_POST['anisave'])) {
$id = $_POST['id'];
$title = $_POST['title'];
$genre = $_POST['genre'];
$description = $_POST['description'];
$start = $_POST['start'];
$stop = $_POST['stop'];
$image_file = $_FILES['image']['name'];
$type = $_FILES['image']['type'];
$size = $_FILES['image']['size'];
if (empty($image_file) || empty($id)) {
echo "Sorry, form is not complete yet!";
header('Location: add.php');
}
else{
$query_id = mysql_query("SELECT * FROM anidata WHERE id = '$id'");
$check = mysql_num_rows($query_id);
if ($check > 0) {
echo "Sorry, Anime ID not available";
header('Location: add.php');
}
else{
if ($type != "image/gif" && $type != "image/jpg" && $type != "image/jpeg" && $type != "image/png") {
echo "Invalid image file, please use JPEG,JPG,PNG or GIF to upload the image."
header('Location: add.php');
}
if ($size > 10000) {
echo "Affordable file is under 10mB."
header('Location: add.php');
}
else{
$upload_directory = 'upload/';
$temp = $upload_directory.$image_file;
if (move_uploaded_file($_FILES['image']['tmp_name'] , $temp)) {
$sql = "INSERT INTO anidata VALUES ('$id', '$title', '$temp', '$genre', '$description','$start', '$stop')";
$query = mysql_query($sql)
if ($query) {
header('Location: view.php');
}
else{
echo mysql_query();
}
}
else{
echo "<p> Upload Failed, error code = " . $_FILES['location']['error']. "</p>";
}
}
}
}
}
else{
unset($_POST['anisave']);
}
?>