Return 不同的行仅基于 oracle 中的一列 sql

Question

我想要 return n 个不同的行。不同的行应仅基于一列 (SN)。

我有下面的查询，预计 return 4 行序列号大于 2 并且没有具有相似 SN 列值的行被 returned。

Table

SN  letter value
 1   test   25
 1   bread  26
 3   alpha  43
 4   beta   23
 4   gamma  5
 5   omega  60
 6   omega  60

预期结果

SN  letter value
 3   alpha  43
 4   beta   23
 5   omega  60
 6   omega 60

这是我的查询。这不能正常工作，它 return 是重复项，因为它通过组合的所有列而不是单个列 SN 过滤不同的值。

SELECT * FROM (SELECT a.*, row_number() over(order by SN) rowRank 
FROM (SELECT distinct SN, letter, value from table where SN > 2 order by SN) a) 
WHERE rowRank BETWEEN 1 AND 4}"

Answer 1

您无需使用 DISTINCT 即可尝试过滤掉您的结果。如果您需要修改应该返回 SN 的哪个副本，您可以修改 row_rank 分析函数的 ORDER BY 子句。现在它按字母顺序返回第一个 LETTER 值，因为它与您的示例结果匹配。

查询

WITH
    some_table (sn, letter, VALUE)
    AS
        (SELECT 1, 'test', 25 FROM DUAL
         UNION ALL
         SELECT 1, 'bread', 26 FROM DUAL
         UNION ALL
         SELECT 3, 'alpha', 43 FROM DUAL
         UNION ALL
         SELECT 4, 'beta', 23 FROM DUAL
         UNION ALL
         SELECT 4, 'gamma', 5 FROM DUAL
         UNION ALL
         SELECT 5, 'omega', 60 FROM DUAL
         UNION ALL
         SELECT 6, 'omega', 60 FROM DUAL)
  --Above is to set up the sample data. Use the query below with your real table
  SELECT sn, letter, VALUE
    FROM (SELECT sn,
                 letter,
                 VALUE,
                 ROW_NUMBER () OVER (PARTITION BY sn ORDER BY letter)     AS row_rank
            FROM some_table
           WHERE sn > 2)
   WHERE row_rank = 1
ORDER BY sn
   FETCH FIRST 4 ROWS ONLY;

结果

   SN    LETTER    VALUE
_____ _________ ________
    3 alpha           43
    4 beta            23
    5 omega           60
    6 omega           60

Answer 2

SELECT * FROM 
    (
        SELECT 
        t.* 
        ,ROW_NUMBER() OVER (PARTITION BY sn ORDER BY value ) rn
        FROM 
        t
        WHERE sn > 2
    ) t1
 WHERE t1.rn = 1
 ORDER BY sn;

Return 不同的行仅基于 oracle 中的一列 sql

Return distinct rows based on only one column in oracle sql

sql

oracle

plsql

查询

结果