在oracle中将一列数据拆分为多列

Question

在我的 oracle 查询中，我使用如下方式检索记录，结果如下所示 -

SELECT columnC
     , LISTAGG(r.columnA,',') WITHIN GROUP (ORDER BY r.columnB) AS Test_sensor
  FROM tableA
 GROUP BY columnC

目前的输出如下 -

ColumnC  |  Test_Sensor
=============================
Z12345   |  20,30,40,50,60,70

但我希望这些数据显示如下 -

ColumnC  |  Test_Sensor1 |  Test_Sensor2 |   Test_Sensor3  |  Test_Sensor4
==========================================================================
Z12345   |  20           |   30          |    40           |  50   

请帮我解决这个问题

谢谢 克兰蒂 RTR

Answer 1

您可以使用PIVOT（并且不需要使用LISTAGG）：

SQL Fiddle

Oracle 11g R2 架构设置：

CREATE TABLE TableA ( ColumnA, ColumnB, ColumnC ) AS
  SELECT 20, 'A', 'Z12345' FROM DUAL UNION ALL
  SELECT 30, 'B', 'Z12345' FROM DUAL UNION ALL
  SELECT 40, 'C', 'Z12345' FROM DUAL UNION ALL
  SELECT 50, 'D', 'Z12345' FROM DUAL UNION ALL
  SELECT 60, 'E', 'Z12345' FROM DUAL UNION ALL
  SELECT 70, 'F', 'Z12345' FROM DUAL;

查询 1：

SELECT *
from (
  SELECT columnA,
         columnC,
         ROW_NUMBER() OVER ( PARTITION BY columnC ORDER BY columnB ) AS rn
  FROM   tableA
) a
PIVOT ( MAX( columnA ) FOR rn IN (
  1 AS test_sensor1,
  2 AS test_sensor2,
  3 AS test_sensor3,
  4 AS test_sensor4
) )

Results：

| COLUMNC | TEST_SENSOR1 | TEST_SENSOR2 | TEST_SENSOR3 | TEST_SENSOR4 |
|---------|--------------|--------------|--------------|--------------|
|  Z12345 |           20 |           30 |           40 |           50 |

查询 2：

您可以使用LISTAGG 来做到这一点，但它比使用PIVOT 效率低得多：

SELECT ColumnC,
       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 1 ) AS test_sensor1,
       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 2 ) AS test_sensor2,
       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 3 ) AS test_sensor3,
       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 4 ) AS test_sensor4
FROM   (
  SELECT ColumnC,
         LISTAGG( ColumnA, ',' ) WITHIN GROUP ( ORDER BY ColumnB )
           AS test_sensor
  FROM   TableA
  GROUP BY ColumnC
)

Results：

| COLUMNC | TEST_SENSOR1 | TEST_SENSOR2 | TEST_SENSOR3 | TEST_SENSOR4 |
|---------|--------------|--------------|--------------|--------------|
|  Z12345 |           20 |           30 |           40 |           50 |

Answer 2

如果您使用 LISTAGG，则需要注意 LISTAGG 会忽略 NULL 值。你确定你的 columnA 总是有数据吗？例如，如果 test_sensor2 为 NULL，则 listagg 操作的输出将为 20,40,50,60,70，因此在 NULL 值之后，所有传感器的数据都将报告错误！使用这个来纠正它：

replace(LISTAGG( nvl(to_char(ColumnA), ','), ',' ) 
        WITHIN GROUP ( ORDER BY ColumnB ), ',,,',',,')

现在你的输出是 20,,40,50,60,70，保持 test_sensor2 的 NULL 值和其余的在正确的位置。

但是现在还有另一个问题是'[^,]+' 形式的正则表达式会导致同样的问题！因此，即使 listagg 输出是固定的，解析后的输出在具有 NULL 值的列之后又回到关闭状态！下面显示了一种不同的形式来解决此问题。这是一个设置示例，因此您可以评论/取消评论以查看处理数据时的差异。总是期待意外！

with TableA ( ColumnA, ColumnB, ColumnC ) AS (
  SELECT 20, 'A', 'Z12345' FROM DUAL UNION ALL
  SELECT NULL, 'B', 'Z12345' FROM DUAL UNION ALL -- make NULL
  SELECT 40, 'C', 'Z12345' FROM DUAL UNION ALL
  SELECT 50, 'D', 'Z12345' FROM DUAL UNION ALL
  SELECT 60, 'E', 'Z12345' FROM DUAL UNION ALL
  SELECT 70, 'F', 'Z12345' FROM DUAL
),
tbl_tmp as (
  SELECT ColumnC,
         -- Preserve the NULL in position 2
         replace(LISTAGG( nvl(to_char(ColumnA), ','), ',' ) 
                 WITHIN GROUP ( ORDER BY ColumnB ), ',,,',',,')
           AS test_sensor
  FROM   TableA
  GROUP BY ColumnC
)
--select * from tbl_tmp;
-- regex of format [^,]+ does not handle NULLs
SELECT ColumnC,
--       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 1) AS test_sensor1,  
--       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 2 ) AS test_sensor2,
--       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 3 ) AS test_sensor3,
--       REGEXP_SUBSTR( test_sensor, '[^,]+', 1, 4 ) AS test_sensor4
       REGEXP_SUBSTR( test_sensor, '(.*?)(,|$)', 1, 1, NULL, 1) AS test_sensor1,
       REGEXP_SUBSTR( test_sensor, '(.*?)(,|$)', 1, 2, NULL, 1 ) AS test_sensor2,
       REGEXP_SUBSTR( test_sensor, '(.*?)(,|$)', 1, 3, NULL, 1 ) AS test_sensor3,
       REGEXP_SUBSTR( test_sensor, '(.*?)(,|$)', 1, 4, NULL, 1 ) AS test_sensor4
FROM  tbl_tmp;