Question

如何列出具有相同姓氏的所有乐队成员的姓名？

该列的值如下

band_NAME
-------------------
Carla Thomas
Stephen E. Rice
Cynthia P. Tree
Richard Anthony Paul
Ann Frances Smith
Lorace Black
Timothy Adam Paul

我知道我们必须使用instr和substr。我只是不知道如何确定这个位置。

我知道基本格式会像

SELECT band_NAME 
FROM TABLE 
where substr(band_name, ?, instr( ) IN 
    (select substr(band_name, ?, instr( )-1)  
     from table 
     group by SUBSTR(band_NAME , ?, INSTR( )-1 ) 
     HAVING COUNT(* ) > 1 );

但是instr中的问号和内部是什么？

非常感谢您的帮助！

Answer 1

我假设您的名字和名字之间的分隔符是单个空格。字符串中的更多空格是姓氏的一部分。因此，您可能希望搜索第一个空格字符。

使用instr(str, substr)返回第一次出现的子串的位置。

然后，使用substring(str, pos)返回从给定位置开始的子字符串（由instr函数提供）。

SELECT substring(band_name, instr(band_name, ' '))
FROM yourtable

Answer 2

试试这个：

SELECT t1.band_NAME
FROM TABLE t1 LEFT JOIN TABLE t2
 ON SUBSTRING_INDEX(t1.band_name, ' ', - 1) = SUBSTRING_INDEX(t2.band_name, ' ', - 1)
WHERE t1.band_name <> t2.band_name

这就像你的伪代码 MySQL ：

SELECT band_NAME FROM TABLE
Where FIND_IN_SET (SUBSTRING_INDEX(band_name, ' ', -1),
 (Select SUBSTRING_INDEX(band_name, ' ', -1) bn
 From TABLE Group by bn
 having Count(bn) > 1
 )
)

SQL Server

SELECT band_NAME FROM TABLE
Where
 SUBSTRING(band_NAME, CHARINDEX(' ', band_NAME) + 1, LEN(band_NAME)) AS [Last Name]
IN
 (Select SUBSTRING(band_NAME, CHARINDEX(' ', band_NAME) + 1, LEN(band_NAME)) AS [Last Name]
 From TABLE Group by [Last Name] -- or SUBSTRING(band_NAME, CHARINDEX(' ', band_NAME) + 1, LEN(band_NAME)) AS [Last Name]
 having Count(*) > 1
 )
)

此外，我认为您可以通过某种方式受益于STRING_SPLIT

Answer 3

试试这个

 with cte as
(
select band_name, ROW_NUMBER() over(partition by SUBSTRING(band_name,CHARINDEX(' ',band_name),LEN(band_name))  order by band_name) as cnt,
        SUBSTRING(band_name,CHARINDEX(' ',band_name),LEN(band_name)) as lastname
from your_table
)
select band_name 
from cte
where lastname in (select lastname from cte where cnt > 1)

如何在列中查找第二个值

3 个答案: