sql - 检测oracle表中所有在每一行中具有相同值的列

Question

每天，请求变得越来越奇怪。

我被要求组合一个查询来检测表中的哪些列为所有行包含相同的值。我说:“这需要通过程序来完成，以便我们可以在表的一遍而不是N遍中进行。”

我被否决了。

长话短说。我有这个非常简单的查询，它说明了问题。它使测试集通过4次。我正在寻找有关SQL Magery的想法，这些想法不涉及在每一列上添加索引，编写程序或占用整个人的生命。

和叹息它需要能够在任何表上工作。

预先感谢您的建议。

WITH TEST_CASE AS
(
    SELECT 'X' A, 5 B, 'FRI' C, NULL D FROM DUAL UNION ALL
    SELECT 'X' A, 3 B, 'FRI' C, NULL D FROM DUAL UNION ALL
    SELECT 'X' A, 7 B, 'TUE' C, NULL D FROM DUAL 
),
KOUNTS AS 
(
    SELECT SQRT(COUNT(*)) S, 'Column A' COLUMNS_WITH_SINGLE_VALUES
    FROM TEST_CASE P, TEST_CASE Q
    WHERE P.A = Q.A OR (P.A IS NULL AND Q.A IS NULL)

    UNION ALL

    SELECT SQRT(COUNT(*)) S, 'Column B' COLUMNS_WITH_SINGLE_VALUES
    FROM TEST_CASE P, TEST_CASE Q
    WHERE P.B = Q.B OR (P.B IS NULL AND Q.B IS NULL)

    UNION ALL

    SELECT SQRT(COUNT(*)) S, 'Column C' COLUMNS_WITH_SINGLE_VALUES
    FROM TEST_CASE P, TEST_CASE Q
    WHERE P.C = Q.C OR (P.C IS NULL AND Q.C IS NULL)

    UNION ALL

    SELECT SQRT(COUNT(*)) S, 'Column D' COLUMNS_WITH_SINGLE_VALUES
    FROM TEST_CASE P, TEST_CASE Q
    WHERE P.D = Q.D OR (P.D IS NULL AND Q.D IS NULL)
)
SELECT COLUMNS_WITH_SINGLE_VALUES
FROM KOUNTS
WHERE S = (SELECT COUNT(*) FROM TEST_CASE)

Answer 1

你的意思是这样的吗？

WITH 
TEST_CASE AS
(
    SELECT 'X' A, 5 B, 'FRI' C, NULL D FROM DUAL UNION ALL
    SELECT 'X' A, 3 B, 'FRI' C, NULL D FROM DUAL UNION ALL
    SELECT 'X' A, 7 B, 'TUE' C, NULL D FROM DUAL 
)
select case when min(A) = max(A) THEN 'A'
            when min(B) = max(B) THEN 'B'
            when min(C) = max(C) THEN 'C'
            when min(D) = max(D) THEN 'D'
            else 'No one'
       end 
from TEST_CASE

编辑
这有效:

WITH 
TEST_CASE AS
(
    SELECT 'X' A, 5 B, 'FRI' C, NULL D FROM DUAL UNION ALL
    SELECT 'X' A, 3 B, 'FRI' C, NULL D FROM DUAL UNION ALL
    SELECT 'X' A, 7 B, 'TUE' C, NULL D FROM DUAL 
)
select case when min(nvl(A,0)) = max(nvl(A,0)) THEN 'A ' end ||
       case when min(nvl(B,0)) = max(nvl(B,0)) THEN 'B ' end ||
       case when min(nvl(C,0)) = max(nvl(C,0)) THEN 'C ' end ||
       case when min(nvl(D,0)) = max(nvl(D,0)) THEN 'D ' end c

from TEST_CASE

奖励:我还添加了对null值的检查，所以现在的结果是:A和D

还有 SQLFiddle demo为您服务。