如果mysql没有产生任何结果，如何给变量赋值？

Question

有两个数据库。我正在尝试查看第一个数据库中的电子邮件是否也存在于第二个数据库中。下面是我尝试的代码:

// Getting info from the first "subscribers" database.
 
$sql = "SELECT name, email, status 
        FROM subscribers 
        WHERE status='active'";

$result = mysqli_query($link,$sql);

while($row = mysqli_fetch_assoc($result)) {

    // To check if all emails in "subscribers" database are also present in "phpbb_users" database. 
    // If they are present then user type=0 otherwise user type=1.

    $subscriber_email = $row['email'];
    $forum_sql = mysqli_query($forum_link, "SELECT user_type 
                                            FROM phpbb_users 
                                            WHERE user_email='$subscriber_email'");

    while($forum_row = mysqli_fetch_assoc($forum_sql)){
        $ut = $forum_row['user_type'];
        if($ut=="0"){
            $type="0";
        } else {
            $type="1";
        }
        echo $type;
    }   

    // To diplay the data from "subscribers" database.
    echo $row["name"]; 
    echo $row['email'];
} 

输出为:

null

Answer 1

试试这条线。。

echo empty($ut) ? 1 : $ut;

Answer 2

null

select name, email, 
    exists (select 1 from phpbb_users u where u.user_email = s.email) as exists_in_phpbb
from subscribers s
where status = 'active'

这将向结果集添加一个名为exists_in_phpbb的列，该列包含电子邮件的1，如果不包含，则可以在phpbb_users中找到0。

Answer 3

这是本节代码中的一个简单逻辑问题:

while($forum_row = mysqli_fetch_assoc($forum_sql)){
    $ut = $forum_row['user_type'];
    if($ut=="0"){
        $type="0";
    } else {
        $type="1";
    }
    echo $type;
}

null

null

$type = "1";
while($forum_row = mysqli_fetch_assoc($forum_sql)) {
    $ut = $forum_row['user_type'];
    if($ut == "0"){
        $type = "0";
    }
}
echo $type;