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Тема |
Re: Въвеждане на данни в 2 таблици??? [re: salle] |
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Автор |
tonchita (непознат
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Публикувано | 27.12.08 16:57 |
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Вижте какво направих, но нещо не работи.
След като попълня формата и натисна Submit ми изписва следното:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
<?php
mysql_connect('localhost', 'root', '') or die('Could not connect: ' . mysql_error());
mysql_select_db('hotel');
///////////////////////////////////////////////////////////////
$username = $_POST['username'];
$password = $_POST['password'];
$datetime = $_POST['datetime'];
$day_validation = $_POST['day_validation'];
$level_user = $_POST['level_user'];
$insertSQL = sprintf("INSERT INTO accounts (account_id, username, password, datetime, day_validation, level_user) VALUES ('', '$username', '$password', '$datetime', '$day_validation', '$level_user'");
$Result = mysql_query($insertSQL) or die(mysql_error());
$name = $_POST['name'];
$second_name = $_POST['second_name'];
$last_name = $_POST['last_name'];
$room = $_POST['room'];
$account = $_POST['account'];
$insert3SQL = sprintf("INSERT INTO info (name, second_name, last_name, room, account) VALUES ('$name', '$second_name', '$last_name', '$room', LAST_INSERT_ID()");
$Result3 = mysql_query($insert3SQL) or die(mysql_error());
?>
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