-3

该脚本应从数据库中选择 StaffID,然后将其放入一个数组中,然后我可以从那里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

'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:未定义索引:E:\EasyPHP-12.1\www\Placing_Orders.php 112 行中的staffID

为什么是这样?

4

1 回答 1

3

http://php.net/manual/en/function.mysql-fetch-array.php

Note: Field names returned by this function are case-sensitive.

于 2013-05-13T11:57:24.207 回答