SQL - 選擇結束了在合併

Question

我嘗試了一些重複的做一個簡單的選擇，當一個錯誤，但我的sql語句總是與這個錯誤結束：

Command not properly ended

我有一個表像這樣

EAN    STR 
============= ==== 
8030524096397 AAAA 
8030524096397 BBBB 
8030524096731 XXXX 
8030524096731 YYYY 
8030524096324 CCCC 

我的選擇其實很簡單

SELECT EAN, COUNT(*) FROM ean GROUP BY ean HAVING COUNT(*) > 1; 

Reults：

EAN    COUNT(*) 
=============  ======== 
8030524096397  2 
8030524096731  2 

一切都很好，直到這裏！現在，我想重複的STR和嘗試這個

SELECT * FROM EAN E 
    INNER JOIN (SELECT EAN, COUNT(*) FROM ean GROUP BY ean HAVING COUNT(*) > 1) 
R ON 
E.EAN = R.EAN; 

但是這將導致這個錯誤。

它正是這樣說：

SELECT * FROM EAN E 
INNER JOIN (SELECT EAN, COUNT(*) FROM ean GROUP BY ean HAVING COUNT(*) > 1) R ON 
^ 
Error: Command not properly ended 

我在做什麼錯？

信息到DB：古普塔的Centura SQLBase 7.0.1

Answer 1

我不認爲SQLBase 7.01支持ANSI正確的語法JOIN（旁白：有什麼好理由使用更現代化的產品）。該錯誤表示INNER JOIN存在問題。

這裏有兩種可能的解決方案。

首先，難吃陳舊join語法：

SELECT * 
FROM EAN E, 
    (SELECT EAN, COUNT(*) as cnt FROM ean GROUP BY ean HAVING COUNT(*) > 1) 
R 
WHERE E.EAN = R.EAN; 

二，IN：

SELECT * 
FROM EAN E 
WHERE E.EAN IN (SELECT EAN FROM ean GROUP BY ean HAVING COUNT(*) > 1) 

Answer 2

SELECT * 
FROM EAN E 
INNER JOIN 
    (SELECT EA.EAN, COUNT(*) Cnt 
    FROM ean EA 
    GROUP BY EA.ean 
    HAVING COUNT(*) > 1 
    ) R ON E.EAN = R.EAN; 

我給了一個別名CNT到COUNT（*）。這應該會讓錯誤消失。

Answer 3

試試這個（添加別名的表與AS關鍵字）

SELECT * FROM EAN AS E 
INNER JOIN 
(SELECT EAN, COUNT(*) FROM ean GROUP BY ean HAVING COUNT(*) > 1) AS R 
ON 
E.EAN = R.EAN;