Question

我有桌子
具有 id，name，check_in_id，check_out_id 的Patient_Detail 。
{Check_Out Check_Out_ID，Illness_ID，Check_Out_Date。

Check_In Check_In_ID，Illness_ID，Check_In_Date。

Illness有 Illness_ID，Illness_Name。

但问题是我不知道如何通过illness_Name表加入Check_out。< / p>

USE PantienDatabase

SELECT Name,Check_in,Illness_name,Check_out_ID FROM Check_In AS CI
INNER JOIN Patient_Detail AS P 
ON CI.Check_In_ID = P.Check_In_ID 
INNER JOIN Illness AS I 
ON I.IllnessID  =CI.illness_ID
INNER JOIN 
Check_Out AS CO
ON CO.Check_Out_ID = P.Check_out_ID

Answer 1

您可以将同一个表连接两次，并为所选列使用别名：

SELECT Name, Check_in,
       II.Illness_name AS Illness_name_in,
       IO.Illness_name AS Illness_name_out,
       Check_out_ID
FROM Check_In AS CI
INNER JOIN Patient_Detail AS P ON CI.Check_In_ID = P.Check_In_ID 
INNER JOIN Illness AS II ON II.IllnessID = CI.illness_ID
INNER JOIN Check_Out AS CO ON CO.Check_Out_ID = P.Check_out_ID
INNER JOIN Illness AS IO ON IO.IllnessID = CO.illness_ID

如何内连接表？

1 个答案: