如果数据不在其中一个表中，则PHP连接三个表

Question

我正在努力将三个表连接在一起，但在其中一个表中没有信息将在那里，除非用户打开我们发送给他们的电子邮件

我目前正在使用这个SQL，但它似乎无法正常工作

SELECT email.senton, customer.*, count(tracker.viewed) as viewed 
FROM email_sent AS email, customer_detail AS customer, email_tracker AS tracker 
WHERE email.customer_id = customer.customer_id 
    AND customer.Email = tracker.emailaddress 
    AND customer.LeadOwnerId = '{$this->userid}'

这是由于表email_tracker可能没有客户信息，除非客户已打开电子邮件

Answer 1

尝试这个：

SELECT email.senton, customer.*, COUNT(tracker.viewed) as viewed 
FROM email_sent email INNER JOIN customer_detail customer
    ON email.customer_id = customer.customer_id
LEFT JOIN email_tracker tracker 
    ON customer.Email = tracker.emailaddress
WHERE customer.LeadOwnerId = '{$this->userid}'

LEFT JOIN子句用于始终获取左侧部分的列和右侧部分的列（如果它们存在于连接中...

根据您的评论编辑：
尝试更改COUNT(tracker.viewed)部分

CASE
    WHEN tracker.emailaddress IS NOT NULL THEN COUNT(tracker.viewed)
    ELSE 0
END
    as viewed 

我不确定它是否有效，我无法测试它，但试一试

Answer 2

您需要的行为是LEFT OUTER JOIN （也称为LEFT JOIN ）。 如果该行在另一个表中不存在，它只将值设置为NULL 。

SELECT email.senton, customer.*, COUNT(tracker.viewed) AS viewed
FROM email_sent AS email
    JOIN customer_detail AS customer USING(customer_id)
    LEFT OUTER JOIN email_tracker AS tracker ON customer.Email = tracker.emailaddress
WHERE customer.LeadOwnerId = '{$this->userid}'

PS：它一般是一个更好的主意，使用JOIN ！而非笛卡尔积的（你做了什么, ）。