Question

我一直在努力让以下问题得以运行：

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND if(24hr = 'yes') user.last_mail >= DATE_SUB (user.last_mail INTERVAL 1 DAY)

基本上，我有3个表，包含用户名（wine_users），订阅（wine_subscriptions）和用户设置（user_settings）

我想要的是选择在我们中选择了NO的用户.24小时或在我们中选择YES.24hr（在我们中存储为ENUM'yes''no'。24小时）并且没有收到e-最近24小时内的邮件（这是作为TIMESTAMP存储在user.lastmail中）

任何帮助都会阻止我通过我的左鼻孔撕掉我自己的大脑！：）

谢谢：）

的Seb

哇！你们真棒！ :)尝试了所有的建议，但他们都返回以下错误：

1064 - 您的SQL语法出错;检查手册对应于您的MySQL服务器版本，以便使用正确的语法在'INTERVAL 1 DAY）附近））限制0,30'在第8行

我猜这与DATE_SUB命令有关，不喜欢user.last_mail字段的TIMESTAMP格式......这可能是正确的，如果是这样，我该如何处理呢？

为了清楚起见，脚本是一个页面的一部分，每隔一小时就会被chron点燃，我已经为我做过了。几个朋友。它检查特易购葡萄酒是否有“神秘葡萄酒包”（因为网上设计总是更好用廉价葡萄酒！）
如果有的话（每周只发生一次或所以它通过电子邮件发送用户数据库中的任何人，他们将24小时设置为否（即产品可用时每小时发送一次电子邮件）或者24小时设置为YES并且没有收到电子邮件。最后24小时
（因此last_mail字段，记录发送最后一封电子邮件时的TIMESTAMP）

非常感谢回复！

Seb：）

Answer 1

看看这是否有效。

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND ( (24hr = 'yes' AND user.last_mail >= DATE_SUB (user.last_mail INTERVAL 1 DAY)) OR 24hr='no')

Answer 2

我从你的问题中假设必须满足以下两个条件中的一个：

一个。在我们中选择NO的用户.24hr

B中。在我们中选择YES的用户.24小时并且在过去24小时内没有收到电子邮件

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND (
     user.receive_mail = 'no'
 OR (user.24hr = 'yes'
AND  DATE_SUB(user.last_mail interval 1 day) > user.last_Mail)
    )

Answer 3

如果我没有读错你的问题，你可以使用联盟：

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND if(24hr = 'yes') user.last_mail >= DATE_SUB (user.last_mail INTERVAL 1 DAY)
 UNION ALL
SELECT user.username, user.email, user.id, user.last_mail, us.24hr
  FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND 24hr = 'No'

Answer 4

我想你可能正在寻找这个：

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND (
  (24hr = 'yes' AND user.last_mail >= DATE_SUB (user.last_mail, INTERVAL 1 DAY))
  OR 24hr = 'no'
)

但我建议您使用JOIN进行重新设置，例如：

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users AS user
JOIN (wine_user_settings AS us, wine_subscriptions AS sub)
ON (us.id = user.id AND us.id = sub.user_id)
WHERE sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND (
  (24hr = 'yes' AND user.last_mail >= DATE_SUB (user.last_mail, INTERVAL 1 DAY))
  OR 24hr = 'no'
)

不是我可以测试它。

Answer 5

您可以使用OR：

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND (24hr = 'no' OR
     (24hr = 'yes' AND user.last_mail < DATE_SUB(NOW(), INTERVAL 1 DAY)))

您可以使用UNION ALL：

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND 24hr = 'yes'
AND user.last_mail < DATE_SUB(NOW(), INTERVAL 1 DAY)

UNION ALL

SELECT user.username, user.email, user.id, user.last_mail, us.24hr
FROM wine_users user, wine_subscriptions sub, wine_user_settings us
WHERE user.id = sub.user_id
AND us.id = user.id
AND sub.wine1 = 'yes'
AND user.receive_mail = 'yes'
AND 24hr = 'no'

您有时会发现UNION ALL更有效率，即使它打字更多，因为数据库引擎可以找到更优化的执行计划。

Answer 6

user.last_mail >= DATE_SUB (user.last_mail INTERVAL 1 DAY)永远不会是False！

我认为你需要使用函数NOW()和一些括号OR：

SELECT user.username
     , user.email
     , user.id
     , user.last_mail
     , us.24hr
FROM wine_users AS user
    JOIN wine_subscriptions AS sub
        ON user.id = sub.user_id
    JOIN wine_user_settings AS us
        ON us.id = user.id
WHERE sub.wine1 = 'yes'
  AND user.receive_mail = 'yes'           --- not sure if this line is needed
                                          --- what does `receive_mail` stand for?
  AND ( user.24hr = 'no'
      OR user.last_mail < NOW() - INTERVAL 1 DAY
      OR user.last_mail IS NULL
      )

MySQL条件选择

6 个答案: