Sveiki.
Tātad es meiģinu dabūt iekš datubāzes uzrakstīto informāciju, bet nekas nesanāk, un nekādu erroru ari nav! :(
Esmu jauniņais @ PHP! :D
Kodi:
form.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" lang="lv">
<head>
<meta charset="utf-8" />
<meta name="keywords" content="huh.lv, help, forum, php, css, html, psd, js. skripti" />
<meta name="description" content="HUH.LV - Help forum, join us" />
<title>HUH.LV - Help forum</title>
<link rel="stylesheet" href="style_css/main_style.css" />
<link rel="stylesheet" href="style_css/menu_style.css" />
</head>
<body>
<form name="post news" action="insert.php" method="post">
Title: <input type="text" name="title" />
Text: <input type="text" name="text" />
<input type="submit" name="Submit" value="SUBMIT" />
</form>
</body>
</html>
insert.php
<?
include('connect.php');
$query = "INSERT INTO news(title, text)VALUES('".$_POST['title']."','".$_POST['text']."', now())";
mysql_query($query);
?>
connect.php
<?php
$connect = mysql_connect('localhost','user','12345') or die(mysql_error());
$select_db = mysql_select_db('home', $connect) or die (mysql_error());
$result = mysql_query('SELECT * FROM news ORDER BY id') or die(mysql_error());
$row = mysql_fetch_array($result) or die(mysql_error());
?>
