Question

我有一张类似

的表格

TRANSACTION_ID  VALUE   TRANSACTION_TYPE    AGR_NBR PART_NBR    
1               3456      BOOKING            ABC1    1    
2               345678    BOOKING            ABC2   NULL    
3               45678     BOOKING            ABC1   2    
4               45678     BIING              ABC2   NULL    
5               45678     BILLING            ABC1   1    
6               4775678   BILLING            ABC2   NULL

我在这张桌子上运行一个简单的选择，如下所示：

SELECT AGR_NBR,PART_NBR,

CASE WHEN TRANSACTION_TYPE = 'BOOKING' 
    THEN SUM(VALUE)
    END AS BOOKING_VAL,

CASE WHEN TRANSACTION_TYPE = 'BILLING' 
    THEN SUM(VALUE)
    END AS BILLINNG_VAL

FROM AA_TEST_TABLE
WHERE AGR_NBR = 'ABC1'

GROUP BY AGR_NBR,PART_NBR,TRANSACTION_TYPE;

来自上述查询的输出是：

AGR_NBR PART_NBR    BOOKING_VAL BILLINNG_VAL    
ABC1    1           NULL        45678    
ABC1    1           3456        NULL    
ABC1    2           45678       NULL

但我想要的是同样的AGR_NBR＆amp;在这种情况下，像（ABC1＆amp; 1）这样的部分应该只有1行，如下所示。如果可能的话，我宁愿做一个单一的选择陈述，因为原始数据有很多公式和＆amp;加入＆amp;这也只是1行的虚拟数据，但实际上会有N行。

AGR_NBR PART_NBR    BOOKING_VAL BILLINNG_VAL    
ABC1    1           3456        45678

Answer 1

试试这个：

SELECT A.AGR_NBR, A.PART_NBR, 
       SUM(CASE WHEN A.TRANSACTION_TYPE = 'BOOKING' THEN A.VALUE ELSE 0 END) AS BOOKING_VAL,
       SUM(CASE WHEN A.TRANSACTION_TYPE = 'BILLING' THEN A.VALUE ELSE 0 END) AS BILLINNG_VAL
FROM AA_TEST_TABLE A
WHERE A.AGR_NBR = 'ABC1'
GROUP BY A.AGR_NBR, A.PART_NBR;

Answer 2

试试这个：

     SELECT AGR_NBR, PART_NBR, SUM(BOOKING_VAL) AS BOOKING_VAL, SUM(BILLING_VAL) AS BILLING_VAL FROM 
       ( 
          SELECT AGR_NBR, PART_NBR, CASE WHEN TRANSACTION_TYPE = 'BOOKING' THEN SUM( VALUE ) END AS BOOKING_VAL, CASE WHEN TRANSACTION_TYPE = 'BILLING' THEN SUM( VALUE ) END AS BILLING_VAL FROM test WHERE AGR_NBR = 'ABC1' GROUP BY AGR_NBR, PART_NBR, TRANSACTION_TYPE 
       )  
     AS t GROUP BY AGR_NBR, PART_NBR

将列连接在一行中

2 个答案: