使用ROWNUM在Oracle中導致無效的標識符錯誤

以下查詢嵌套子查詢中選擇記錄給我無效的標識符錯誤（a.id），因爲它裏面嵌套子查詢：使用ROWNUM在Oracle中導致無效的標識符錯誤

SELECT a.*, 
CASE WHEN 
SELECT id FROM (SELECT id, ROWNUM rnum FROM US b WHERE b.id = a.id ORDER BY b.createdate ASC) WHERE rnum = 2) = 21 THEN ‘Found’ 
END SEARCH 
FROM EU a 
JOIN US b ON b.id = a.id;

任何人都可以給我建議另一種方法？

來源

2015-06-21 Adds

你沒有描述你的設置或你想達到什麼，但這會解決嗎？

SQL Fiddle

的Oracle 11g R2架構設置：

CREATE TABLE EU (id) AS 
SELECT 19+LEVEL 
FROM DUAL CONNECT BY LEVEL <= 5; 

CREATE TABLE US (id, createdate) AS 
SELECT 19+LEVEL, SYSDATE - LEVEL 
FROM DUAL CONNECT BY LEVEL <= 5 
UNION ALL 
SELECT 19+2*LEVEL, SYSDATE-LEVEL-5 
FROM DUAL CONNECT BY LEVEL <= 3;

查詢1：

SELECT a.*, 
     CASE WHEN a.id = 21 
      AND ROW_NUMBER() OVER (PARTITION BY a.id ORDER BY b.createdate) = 2 
      THEN 'Found' 
      END AS SEARCH 
FROM EU a 
JOIN US b 
ON  b.id = a.id

Results：

| ID | SEARCH | 
|----|--------| 
| 20 | (null) | 
| 21 | (null) | 
| 21 | Found | 
| 22 | (null) | 
| 23 | (null) | 
| 23 | (null) | 
| 24 | (null) |

來源

2015-06-21 22:39:47 MT0

您的查詢的一個問題是在第二個SELECT之前沒有paren。我仍然認爲它不會起作用，因爲Oracle將標識符的範圍限制在一層以內。但是，這是我認爲你打算：

SELECT a.*, 
     (CASE WHEN (SELECT id 
        FROM (SELECT id, ROWNUM as rnum 
         FROM US b 
         WHERE b.id = a.id 
         ORDER BY b.createdate ASC 
         ) 
        WHERE rnum = 2 
       ) = 21 THEN ‘Found’ 
     END) SEARCH 
FROM EU a JOIN 
    US b 
    ON b.id = a.id;

最簡單的方式得到你想要的是使用row_number()：因爲id用於一切

select a.*, 
     (case when b.id = 21 then 'Found' end) as Search 
from eu a left join 
    (select b.*, 
      row_number() over (partition by b.id order by b.createddate) as seqnum 
     from us b 
    ) b 
    on b.id = a.id 
where seqnum = 2;

查詢似乎很奇怪。您好像在查找ID爲21的兩條記錄。

來源

2015-06-21 23:01:31

使用ROWNUM在Oracle中導致無效的標識符錯誤

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