1
我有一个问题我的脚本有三个mysql_query
应后相互使用,我想创建一个脚本,预订门票通过改变他们的身份售出=“无”到“有” ,该脚本计算用户在html表单上输入的票数,该表单向服务器端提供了一个变量,其中包含票数= $tickets
。PHP的MySQL fetch语句获取问题
提示:这是这样一个模型,以便无需MySQL的注射安全
这里是我的代码:
//get ticket status
$eventTicket = mysql_query("SELECT eventTickets FROM beventreservation WHERE eventId = '$eventId'") or die(mysql_error());
$ticketrow = mysql_fetch_array($eventTicket) or die(mysql_error());
//test... which is working !
echo $ticketrow['eventTickets'];
//get classId from classes
$selectClass = mysql_query("SELECT classId FROM quotaclasses WHERE className = '$classes' AND eventFK = '$eventId'") or die (mysql_error());
$classrow = mysql_fetch_array($selectClass) or die(mysql_error());
//this var is to define which class the user used
$choosedClass = $classrow['classId'];
//test ... which did not work !!!
echo $classrow['classId'];
if ($ticketrow['eventTickets'] == "Yes")
{
for($counter=1;$counter<$numberOfTickets;$counter++)
{
$bookTicket = mysql_query("UPDATE unites SET ticketSold = 'Yes' WHERE businessreservationIdFk = '$eventId' AND classIDfk ='$choosedClass'") or die(mysql_error());
echo "ticket ". $counter . " done !";
}
}
脚本不取此语法,并没有错误显示在我的页面上!
$classrow = mysql_fetch_array($selectClass) or die(mysql_error());
还,我想这句法后呼应变量$tickets
,它没有露面,有没有要读取的mysql_query多个相同的脚本页面上的问题吗?请告诉我,我在哪里错了。