[英]Generate a rownumber and repeat every five customers in oracle
如何在oracle中為每5個客戶生成一個行號? 例如,如果記錄數= 10,則預期輸出將是: 1..5,1..5
像這樣:
CustomerName Row Number
A 1
B 2
C 3
D 4
E 5
F 1
G 2
H 3
I 4
J 5
一種解決方案是NTILE
分析功能: https : //docs.oracle.com/database/121/SQLRF/functions127.htm#SQLRF00680
它也很好地處理了客戶數不能被5整除的情況(例如12個客戶)。
樣品:
with customers as (
select level customername from dual connect by level <= 10)
select customername, ntile(5) over (order by customername asc) rownumber
from customers;
您可以使用模塊化算術邏輯,如下所示:
select "Customer Name", replace(mod(rn,5),0,5) "Row Number"
from
(
select CustomerName as "Customer Name", row_number() over (order by CustomerName) as rn
from
(
select chr(level+64) CustomerName, level as nr
from dual
connect by level <= 10
)
);
在Oracle中,您可以執行以下操作:
select t.*, 1 + mod(rownum - 1, 5) as rownumber
from t;
您也可以將rownum
替換為row_number() over (order by . . . )
。
這也適用(oracle); 根據您的需要更改
選擇
'data'data_column,
rownum original_rownum,
replace(mod(rownum,5),0,5)Expected_rownum
從
your_table;
我的解決方案使用ROW_NUMBER為每行分配一個值,然后應用MOD函數將其拆分為5。 盡管這可行,但我認為使用NTILE的其他解決方案更干凈。
WITH cust AS
(SELECT customername, ROW_NUMBER() OVER(ORDER BY customer_name) AS ordering
FROM customers)
SELECT customername , CASE WHEN MOD(ordering,5) = 0 THEN 5 ELSE MOD(ordering,5) END AS bucket
FROM cust;
這將做:
select e.*, mod(rownum-1,5)+1 rownumber
from (select * from table_name) e;
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