Код:
<form action="insert.php" method="post">
Heigh<input type="text" name="heigh"><br>
Title <input type="text" name="title">
bold
<input type="text" name="bold" value="1">
italic
<input type="text" name="italic" value="1"><br>
Size <input type="text" name="size"><br>
Picture Url <input type="text" name="url"><br>
Description <input type="text" name="desc"><br>
Size Description <input type="text" name="sized"><br>
Address <input type="text" name="address"><br>
Phone: <input type="text" name="tel"><br>
Fax: <input type="text" name="fax"><br>
E-mail: <input type="text" name="email"><br>
Category <input type="text" name="web"><br>
<input type="Submit">
</form>
Код:
<?
$user="root";
$password="";
$database="kvart";
$heigh=$HTTP_POST_VARS['heigh'];
$title=$HTTP_POST_VARS['title'];
$size=$HTTP_POST_VARS['size'];
$bold=$HTTP_POST_VARS['bold'];
$italic=$HTTP_POST_VARS['italic'];
$url=$HTTP_POST_VARS['url'];
$descr=$HTTP_POST_VARS['descr'];
$sized=$HTTP_POST_VARS['sized'];
$address=$HTTP_POST_VARS['address'];
$tel=$HTTP_POST_VARS['tel'];
$fax=$HTTP_POST_VARS['fax'];
$email=$HTTP_POST_VARS['email'];
$cat=$HTTP_POST_VARS['cat'];
mysql_connect(localhost,$user,$password);
mysql_select_db($database) or die( "DB error (table)");
$query = "INSERT INTO slatina VALUES ('$heigh','$title','$size','$bold','$italic,'$url','$descr','$sized','address','$tel','$fax','$email,'$cat')";
mysql_query($query);
?>
къде е проблемът в този код? защото неможе да се качи в SQl базата данни?