通过使用分析功能组Oracle

Question

我有以下演示代码。 我怎样才能得到想要的output.Its只是没有在我的脑海未来:-(它必须是一个简单的变化

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; 

OUTPUT：。

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;