2

我正在尝试构建一个查询，以便从以前的匹配行构建一些列。例如以下数据：

CREATE TABLE TEST (SEQ NUMBER, LVL NUMBER, DESCR VARCHAR2(10));
INSERT INTO TEST VALUES (1, 1, 'ONE');
INSERT INTO TEST VALUES (2, 2, 'TWO1');
INSERT INTO TEST VALUES (3, 2, 'TWO2');
INSERT INTO TEST VALUES (4, 3, 'THREE1');
INSERT INTO TEST VALUES (5, 2, 'TWO3');
INSERT INTO TEST VALUES (6, 3, 'THREE2');
COMMIT

我想要检索以下数据。 

SEQ L1  L2   L3
1   ONE NULL NULL
2   ONE TWO1 NULL
3   ONE TWO2 NULL
4   ONE TWO2 THREE1
5   ONE TWO3 THREE1
5   ONE TWO3 THREE2

即对于第 3 行，它本身具有 L2 的值，对于 L1，它必须转到包含 L1 数据的最新行，在这种情况下是第一行。

我曾尝试查看分析和连接子句，但无法理解解决方案。
有任何想法吗？

4

3 回答 3

5

更新：有一个比我的第一个答案更简单的解决方案。它更具可读性和更优雅，因此我将它放在首位（通常，感谢Tom Kyte）：

SQL> SELECT seq,
  2         last_value(CASE
  3                       WHEN lvl = 1 THEN
  4                        descr
  5                    END IGNORE NULLS) over(ORDER BY seq) L1,
  6         last_value(CASE
  7                       WHEN lvl = 2 THEN
  8                        descr
  9                    END IGNORE NULLS) over(ORDER BY seq) L2,
 10         last_value(CASE
 11                       WHEN lvl = 3 THEN
 12                        descr
 13                    END IGNORE NULLS) over(ORDER BY seq) L3
 14    FROM TEST;

       SEQ L1         L2         L3
---------- ---------- ---------- ----------
         1 ONE                   
         2 ONE        TWO1       
         3 ONE        TWO2       
         4 ONE        TWO2       THREE1
         5 ONE        TWO3       THREE1
         6 ONE        TWO3       THREE2

以下是我的初步解决方案：

SQL> SELECT seq,
  2         MAX(L1) over(PARTITION BY grp1) L1,
  3         MAX(L2) over(PARTITION BY grp2) L2,
  4         MAX(L3) over(PARTITION BY grp3) L3
  5    FROM (SELECT seq,
  6                 L1, MAX(grp1) over(ORDER BY seq) grp1,
  7                 L2, MAX(grp2) over(ORDER BY seq) grp2,
  8                 L3, MAX(grp3) over(ORDER BY seq) grp3
  9             FROM (SELECT seq,
 10                          CASE WHEN lvl = 1 THEN descr END L1,
 11                          CASE WHEN lvl = 1 AND descr IS NOT NULL THEN ROWNUM END grp1,
 12                          CASE WHEN lvl = 2 THEN descr END L2,
 13                          CASE WHEN lvl = 2 AND descr IS NOT NULL THEN ROWNUM END grp2,
 14                          CASE WHEN lvl = 3 THEN descr END L3,
 15                          CASE WHEN lvl = 3 AND descr IS NOT NULL THEN ROWNUM END grp3
 16                     FROM test))
 17   ORDER BY seq;

       SEQ L1         L2         L3
---------- ---------- ---------- ----------
         1 ONE                   
         2 ONE        TWO1       
         3 ONE        TWO2       
         4 ONE        TWO2       THREE1
         5 ONE        TWO3       THREE1
         6 ONE        TWO3       THREE2
于 2009-06-10T07:20:15.783 回答
0

您是否只有 3 个级别（或固定数量的级别）？

如果是这样，你可以使用类似的东西，这是非常低效的，但我相信它有效（我不能从这台计算机上运行它，所以它是一个可能需要稍微改动的“盲”代码）：

SELECT COUNTER.SEQ AS SEQ, A.DESCR AS L1, B.DESCR AS L2, C.DESCR AS L3
FROM TABLE AS COUNTER, TABLE AS A, TABLE AS B, TABLE AS C
WHERE
A.SEQ = 
    (SELECT MAX(D.SEQ) FROM TABLE AS D 
     WHERE D.LVL = 1 AND D.SEQ <= COUNTER.SEQ) AND 
B.SEQ = 
    (SELECT MAX(D.SEQ) FROM TABLE AS D 
     WHERE D.LVL = 2 AND D.SEQ <= COUNTER.SEQ) AND
C.SEQ = 
    (SELECT MAX(D.SEQ) FROM TABLE AS D 
     WHERE D.LVL = 3 AND D.SEQ <= COUNTER.SEQ)
于 2009-06-10T04:58:51.283 回答
0

要使用 connect 子句，您实际上需要一些行之间的链接。所以应该有一些列可以链接到具有所需值的前一行。对于这些字段，很难进行一次选择，因为您需要为每一行设置 2 个子选择来检查其他级别。

如果合适，我会使用 pl/sql 程序。

declare
    cursor c_cur is
    select * from test order by seq asc;

lvl1 test.descr%type := null;
lvl2 test.descr%type := null;
lvl3 test.descr%type := null;

begin

for rec in c_cur loop
    if rec.lvl = 1 then
        lvl1 := rec.descr;
    elsif rec.lvl = 2 then
        lvl2 := rec.descr;
    elsif rec.lvl = 3 then
        lvl3 := rec.descr;
    end if;
    dbms_output.put_line(rec.seq||','||nvl(lvl1, 'null')||','||nvl(lvl2, 'null')||','||nvl(lvl3, 'null'));  
end loop;

end;
/
于 2009-06-10T06:16:27.020 回答