此脚本应从数据库中选择StaffID,然后将其放入一个数组中,然后从那里我可以将其放入一个变量中,以便在下一个查询中使用。
$query = "SELECT StaffID FROM staff WHERE First_Name = '$db_staff_member_first_name' AND Last_Name = '$db_staff_member_last_name'";
$result = mysql_query($query)
or die(mysql_error());
echo $result;
echo $query + "<br>";
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
{
$db_staffid = $row['staffID'];
}
脚本运行后我得到了
'Resource id #4SELECT CustomerID FROM customer WHERE Customer_First_Name = 'Christopher' AND Customer_Last_Name = 'Bennett'Resource id #50'
然后是以下通知:
注意:未定义的索引:staffID in 第112行的E:\ EasyPHP-12.1 \ www \ Placing_Orders.php
为什么会这样?
答案 0 :(得分:3)
http://php.net/manual/en/function.mysql-fetch-array.php
Note: Field names returned by this function are case-sensitive.