我试图从一个表中显示两组数据。我正在为自己的游戏创建一个商店,并拥有一个数据库,其中列出了商店中的pokemon/价格/类型/id。现在我已经让它运行起来了,它会显示商店中的所有pokemon,并且它们下方都有一个购买按钮,但不管你想购买什么pokemon,它都只会购买位于列表顶部的pokemon。我希望我解释得足够好,这是我的代码。
if ($_POST['A'] == '1' ) {
$token= mysql_real_escape_string($_POST['token']);
$tokenn = strip_tags($token);
$sql234 = "SELECT * FROM ticketshop";
$result2 = mysql_query("SELECT * FROM ticketshop");
while($row2 = mysql_fetch_array($result2)){
$sql23 = "SELECT * FROM users WHERE username='".$_SESSION['username']."')";
$result = mysql_query("SELECT * FROM users WHERE username='".$_SESSION['username']."'");
while($row = mysql_fetch_array($result)){
echo "You have ".$row['ticket']." Tickets" ;
echo "<p></p>" ;
if (isset($_POST['slot1'])) {
if ($row['ticket'] >= $row2['price']) {
echo "You have bought ".$row2['pokemon']."" ;
mysql_query("UPDATE users SET ticket=ticket-".$row2['price']." WHERE username='".$_SESSION['username']."'")
or die(mysql_error());
mysql_query("INSERT INTO user_pokemon
(pokemon, belongsto, exp, time_stamp, slot, level, type) VALUES ('".$row2['pokemon']."','".$_SESSION['username']."', 100,'".time()."','0', '5', '".$row2['type']."' )")
or die(mysql_error());
} else {
echo "You can't afford ".$row2['pokemon']."";
}
}
}
}
您正在获得错误,因为您正在关闭与数据库的连接,并且在您尝试从第二个查询中获取结果后,最终出现错误
<?php } /* mysql_close($conn);*/ /*comment or remove closing connection*/ ?>
<p><b>Listing of pending applications</b></p>
然后我想记住你,mysql_
函数是不赞成的,所以我建议你切换到mysqli
或PDO
。