Sql Oracle将外部查询值传递给内部查询_Sql_Oracle

Sql Oracle将外部查询值传递给内部查询

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Sql Oracle将外部查询值传递给内部查询,sql,oracle,Sql,Oracle,这可能是一个简单的方法，但我需要在其他方面应用逻辑： WITH t(col) AS ( SELECT 1 FROM dual UNION SELECT 2 FROM dual UNION SELECT 3 FROM dual UNION SELECT 4 FROM dual UNION SELECT 5 FROM dual ) SELECT col , --- will works as usual (SELECT col FROM t WHERE col = outer_q.col) new

这可能是一个简单的方法，但我需要在其他方面应用逻辑：

WITH t(col) AS (
SELECT 1 FROM dual
UNION SELECT 2 FROM dual
UNION SELECT 3 FROM dual
UNION SELECT 4 FROM dual
UNION SELECT 5 FROM dual
)
SELECT col , --- will works as usual
(SELECT col FROM t WHERE col  = outer_q.col) new_col,  --working as well
(
SELECT sum (latest_col)
from
(
SELECT col latest_col FROM t  WHERE col  = outer_q.col 
UNION ALL
SELECT col FROM t WHERE col   = outer_q.col
)
)newest_col   -- need to get an output "4"
from t outer_q where col  = 2;

一个简单的输出，如：

       COL    NEW_COL NEWEST_COL
---------- ---------- ----------
         2          2          4

我只需要使用第三列中使用的最外面的值和最里面的值

编辑——包含更多数据的示例：

WITH 
t(col) AS
     ( SELECT 1 FROM dual
     UNION
     SELECT 2 FROM dual
     UNION
     SELECT 3 FROM dual
     UNION
     SELECT 4 FROM dual
     UNION
     SELECT 5 FROM dual
     ),
t1(amount, col) AS
     (SELECT 100 , 2 FROM dual
     UNION
     SELECT 200, 3 FROM dual
     )
SELECT col,
     (SELECT col FROM t WHERE col = outer_q.col
     ) new_col,
     (SELECT SUM(x)
     FROM
          (SELECT col x FROM t
          UNION ALL
          SELECT amount x FROM t1
          )
     WHERE col = outer_q.col
     ) newest_col -- gives 315 as it takes whole `SUM`
FROM t outer_q
WHERE col = 2;

预期的输出如下所示：

       COL    NEW_COL NEWEST_COL
---------- ---------- ----------
         2          2        102

提前感谢您的帮助。

内部查询失败，因为您试图将外部的q.col引用向下推两个级别。相关查询只向下一级

Reference:

内部查询失败，因为您试图将外部列引用向下推两级。相关查询只向下一级

参考资料：

好吧，如果您重构了一个查询，那么您可以：

WITH t(col) AS (
  SELECT 1 FROM dual
  UNION SELECT 2 FROM dual
  UNION SELECT 3 FROM dual
  UNION SELECT 4 FROM dual
  UNION SELECT 5 FROM dual
)
SELECT col,
       (SELECT col FROM t WHERE col  = outer_q.col) new_col,
       (SELECT sum (latest_col)
        from
        (
          SELECT col latest_col FROM t 
          UNION ALL
          SELECT col FROM t
        ) x
        where x.latest_col = outer_q.col
       ) newest_col   -- need to get an output "4"
from t outer_q where col = 2;

这在这里是可能的，因为

outer\u q

现在位于子查询的

where

子句中。它以前在子查询（带有

UNION ALL

的子查询）中使用过，而这个子查询正在隐藏它

为了让事情变得更清楚，现在我们有了如下内容：

with t as (...)
select col,
       (SELECT col FROM t WHERE col  = outer_q.col) new_col,
       (SELECT col FROM (Something more complex) WHERE ... = outer_q.col) new_col,
from t outer_q where col = 2;

所以我们现在有了同样的“内部性”

编辑：要回答更新后的问题，需要进行一些调整：

WITH t(col) AS
(
  SELECT 1 FROM dual
  UNION
  SELECT 2 FROM dual
  UNION
  SELECT 3 FROM dual
  UNION
  SELECT 4 FROM dual
  UNION
  SELECT 5 FROM dual
),
t1(amount, col) AS
(
  SELECT 100, 2 FROM dual
  UNION
  SELECT 200, 3 FROM dual
)
SELECT col,
     (SELECT col FROM t WHERE col = outer_q.col) new_col,
     (SELECT SUM(amount)
      FROM
          (SELECT col, col amount FROM t  -- row is (1, 1), then (2, 2) etc
           UNION ALL
           SELECT col, amount FROM t1     -- row is (2, 100), then (3, 200) etc
          )
      WHERE col = outer_q.col
     ) newest_col -- gives 102 as it takes whole `SUM`
FROM t outer_q
WHERE col = 2;

要理解的部分在最里面的查询中：您希望对列和金额值求和，因此重复

col

值，就像它是一个金额一样

获得相同结果的另一种方法（我想性能更好）是将

col

和

amount

加在同一行上：

WITH t(col) AS
(
  SELECT 1 FROM dual
  UNION
  SELECT 2 FROM dual
  UNION
  SELECT 3 FROM dual
  UNION
  SELECT 4 FROM dual
  UNION
  SELECT 5 FROM dual
),
t1(amount, col) AS
(
  SELECT 100, 2 FROM dual
  UNION
  SELECT 200, 3 FROM dual
)
SELECT col,
     (SELECT col FROM t WHERE col = outer_q.col) new_col,
     (SELECT SUM(all_amount)
      FROM
          (SELECT col, col + amount all_amount FROM t1)
      WHERE col = outer_q.col
     ) newest_col -- gives 315 as it takes whole `SUM`
FROM t outer_q
WHERE col = 2;

如果重构一个查询，那么您可以：

WITH t(col) AS (
  SELECT 1 FROM dual
  UNION SELECT 2 FROM dual
  UNION SELECT 3 FROM dual
  UNION SELECT 4 FROM dual
  UNION SELECT 5 FROM dual
)
SELECT col,
       (SELECT col FROM t WHERE col  = outer_q.col) new_col,
       (SELECT sum (latest_col)
        from
        (
          SELECT col latest_col FROM t 
          UNION ALL
          SELECT col FROM t
        ) x
        where x.latest_col = outer_q.col
       ) newest_col   -- need to get an output "4"
from t outer_q where col = 2;