如何计算Oracle数据库中具有相同起始字母的行数

Question

我有一个表客户：

╔════════════╦═════╗
║ CustomerID ║ ... ║
╠════════════╬═════╣
║ ACC20011   ║ ... ║
║ ACC30122   ║ ... ║
║ ACC12356   ║ ... ║
║ EVG12345   ║ ... ║
║ DAA23445   ║ ... ║
║ DAA23548   ║ ... ║
╚════════════╩═════╝

我需要以下输出：

╔════════════╦═══════╗
║ CustomerID ║ Count ║
╠════════════╬═══════╣
║    Acc     ║  3    ║
║    EVG     ║  1    ║
║    DAA     ║  2    ║
╚════════════╩═══════╝

我写了以下查询。但我知道错了。请指正。

select CustomerID , count(*) as Count 
from Customers
where CustomerID is like 'ACC%'

Answer 1

尝试此查询：

SELECT SUBSTR(CustomerID, 1, 3) AS CustomerID, COUNT(*)
FROM Customers
GROUP BY SUBSTR(CustomerID, 1, 3)

Answer 2

如果你有'A1234','ABCC2'等数据格式

，下面的查询也会有效

  SELECT substr(CustomerID, 1, regexp_instr(CustomerID, '\d+') - 1) AS string
        ,count(*) as cnt
    FROM customer
    GROUP BY substr(CustomerID, 1, regexp_instr(CustomerID, '\d+') - 1)

修改 另外一种方法是使用regexp_replace

select regexp_replace(CustomerID,'\d+','') as string,count(*) cnt from customer
group by regexp_replace(CustomerID,'\d+','')

上面一个替换了customerid中的所有数字。