SQL Join 给出 30 个相同的结果答案

【问题标题】：SQL Join gives 30 same resultsSQL Join 给出 30 个相同的结果
【发布时间】：2015-03-08 10:49:38
【问题描述】：

我正在使用连接从两个表中获取消息

表格消息： ID、发件人、收件人、消息、密码

Table OwnMessages： ID、发件人、收件人、消息、密码

查询：

SELECT DISTINCT(Messages.ID) as F_id, 
               Messages.Message as F_message, 
               Messages.Password as F_password, 
               Messages.Receiver as F_receiver, 
               OwnMessages.ID as O_id, 
               OwnMessages.Message as O_message, 
               OwnMessages.Password as O_password, 
               OwnMessages.Sender as O_sender
FROM Messages
JOIN OwnMessages on Messages.Receiver = OwnMessages.Sender
WHERE Messages.Receiver = 8 AND OwnMessages.Sender = 8

Messages 表中的所有内容都正确显示，但从 OwnMessages 表中我得到了第一行 30 次。有什么想法吗？

【问题讨论】：

distinct 是 NOT 一个函数。它适用于select 列表的所有列。
我也试过没有不同的，得到了完全相同的结果
也许 OwnMessages 中有不同值的列（此处未选择）？
如果您添加一些示例表数据和当前结果和想要的结果，帮助您会容易得多！
我建议您删除此问题并提出另一个问题。从样本数据和您想要达到的结果开始。您的查询工作并且完全按照它的设计目的进行。你只是没有表达你真正想要做什么。无效的查询通常不会传达该信息。

标签： mysql sql join

【解决方案1】：

SELECT DISTINCT 
               Messages.ID as F_id, 
               Messages.Message as F_message, 
               Messages.Password as F_password, 
               Messages.Receiver as F_receiver, 
               OwnMessages.ID as O_id, 
               OwnMessages.Message as O_message, 
               OwnMessages.Password as O_password, 
               OwnMessages.Sender as O_sender
FROM Messages
INNER JOIN OwnMessages on Messages.Receiver = OwnMessages.Sender
WHERE Messages.Receiver = 8 AND OwnMessages.Sender = 8

或

SELECT             Messages.ID as F_id, 
                   Messages.Message as F_message, 
                   Messages.Password as F_password, 
                   Messages.Receiver as F_receiver, 
                   OwnMessages.ID as O_id, 
                   OwnMessages.Message as O_message, 
                   OwnMessages.Password as O_password, 
                   OwnMessages.Sender as O_sender
    FROM Messages
    INNER JOIN OwnMessages on Messages.Receiver = OwnMessages.Sender
    WHERE Messages.Receiver = 8 AND OwnMessages.Sender = 8
    GROUP BY Messages.ID
    HAVING COUNT(*) = 1

【讨论】：