Question

这可能很简单，但我需要在其他逻辑中应用逻辑：

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;

简单的输出，如：

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

我只需要将最外面的值用于我用于第三列的内部

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;

预计输出如下：

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

提前感谢您的帮助。

Answer 1

内部查询失败，因为您尝试将outer_q.col引用向下推送两个级别。相关查询仅下降1级

参考：http://asktom.oracle.com/pls/asktom/f?p=100:11:0::::P11_QUESTION_ID:1853075500346799932

Answer 2

好吧，如果你重构一个但是你的查询，你可以这样做：

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;

这是可能的，因为outer_q现在位于子查询的where子句中。之前在子查询中使用它（带有UNION ALL的那个），而这个查询隐藏了它。

为了让事情更清楚，现在我们有类似的事情：

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;

所以我们现在拥有相同水平的“内在性”。

编辑：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;

要理解的部分位于最里面的查询中：您要将列和金额值相加，因此您重复col值，就像它是一个数量一样。

获得相同结果的另一种方法（我猜测性能更高）是在同一行上加col和amount：

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;

Oracle将外部查询值传递给内部查询

2 个答案: