Question

我有以下演示代码。我怎样才能获得所需的输出。它只是没有出现在我的脑海里:-(。它必须是一个简单的改变。

WITH Abc  AS (SELECT 'AB' AS code,'Z12' AS des, 123 AS tk FROM dual UNION
SELECT 'AB','Z14',234 FROM dual UNION
SELECT 'ACB','D12',34 FROM dual UNION
SELECT 'ACB','D12',36 FROM dual UNION
SELECT 'ACB','F12',35 FROM dual UNION
SELECT 'ABX','G44',243 FROM dual UNION
SELECT 'ABX','H44',654 FROM dual UNION
SELECT 'ABX','J12',534 FROM dual UNION
SELECT 'ABQ',NULL,356 FROM dual)

SELECT DISTINCT tk,code,des ,code||'_'||row_number () OVER (PARTITION BY code ORDER BY code,des ) AS str FROM abc
ORDER BY 2,3;

输出：

 tk code des str
--- --- --- ---
123 AB  Z12 AB_1
234 AB  Z14 AB_2
356 ABQ     ABQ_1
243 ABX G44 ABX_1
654 ABX H44 ABX_2
534 ABX J12 ABX_3
34  ACB D12 ACB_1  ---  ACB_1
36  ACB D12 ACB_2  ----I want ACB_1 here too since desc is same for that code
35  ACB F12 ACB_3

Answer 1

**USE RANK() instead of rownumber()**


WITH Abc  AS (SELECT 'AB' AS code,'Z12' AS des, 123 AS tk FROM dual UNION
SELECT 'AB','Z14',234 FROM dual UNION
SELECT 'ACB','D12',34 FROM dual UNION
SELECT 'ACB','D12',36 FROM dual UNION
SELECT 'ACB','F12',35 FROM dual UNION
SELECT 'ABX','G44',243 FROM dual UNION
SELECT 'ABX','H44',654 FROM dual UNION
SELECT 'ABX','J12',534 FROM dual UNION
SELECT 'ABQ',NULL,356 FROM DUAL)

SELECT DISTINCT TK,CODE,DES ,CODE||'_'||rank() OVER (PARTITION BY CODE ORDER BY CODE,DES ) AS STR FROM ABC
ORDER BY 2,3;

通过使用分析功能Oracle进行分组

1 个答案: