Вот текст ошибки:
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 '0 = 99 where id=21' at line 1
[b]rashod.php[/b]
<html>
<?php
session_start();
$db = mysql_connect ("");
mysql_select_db ("");
echo"<form name='update' action='update.php' method='POST'>
<select name='update'>
<option value=''>-- Я --</option>;
<option value='Almaz'>-- Алмаз --</option>;
<option value='Linar'>-- Линар --</option>;
</select></form>";
echo "<form name='update2' action='update.php' method='POST'>
<select name='update2'>
<option value=''>-- Выбери товар для показа --</option>;
<option value='0'>Все</option>";
$result = mysql_query ("SELECT * FROM Tovar")
or die ("<b>Query failed:</b> " . mysql_error());
while ($row = mysql_fetch_array($result)){
echo "<option value=' ".$row['id']." '>".$row['name']."</option>";
}
if ($_SERVER['REQUEST_METHOD'] == "POST"){
$op = $_POST['name'];
mysql_query ("SELECT * FROM Tovar where id=".$op);
}
echo"<br/>";
echo "
<form name='kolvo' action='update.php' method='POST'>
<label><input type='text' name='kolvo' />Количество</label><br/>
<input type='submit' name='Показать' value='Обновить'/><br/>
</form>";
echo "</select></form>";
?>
</html>
update.php
<html>
<?php
session_start();
$db = mysql_connect ("");
mysql_select_db ("");
$strSQL = "UPDATE Tovar SET " . (int)$_POST['update']. " = " . (int)$_POST['kolvo'] . " where id=" . (int)$_POST['update2'] . "";
mysql_query($strSQL) or die (mysql_error());
mysql_close();
?>
<h1>БД обновлена!</h1>
</html>