如何查询匹配特定字符串的行？

Question

这是表格的外观，此处的Empid和idnumber对于每个员工都是唯一的，并且一个员工可以拥有多个具有不同证件编号的证件。

现在，我想过滤以6542和3214开头的徽章的员工，即同时带有6542和3214的徽章的员工

谢谢

更新1

有些记录从6542或3214开始只有一个徽章。但是我只需要带有两个徽章的员工。

Answer 1

就像在徽章字段上一样使用

    Select empid, name 
    from TableName 
    where badge like '6542%' or badge like '3214%'
    group by empid, name
    having count(*)>1

Answer 2

执行GROUP BY，使用HAVING确保两个徽章：

select empid, name
from Table
where badge like '6542-%' or badge like '3214-%'
group by empid, name
having count(distinct badge) > 1

或使用INTERSECT：

select empid, name from Table where badge like '6542-%'
intersect
select empid, name from Table where badge like '3214-%'

Answer 3

这将起作用：

select empid from table_name where regexp_like(Badge,'^(6542)(.*)$')
intersect
select empid from table_name where regexp_like(Badge,'^(3214)(.*)$');

等效的SQL Server：

select empid from table_name where PATINDEX ('^(6542)(.*)$',Badge)  !=0
intersect
select empid from table_name where PATINDEX ('^(6542)(.*)$',Badge)  !=0

Answer 4

我们可以将COUNT DISTINCT中的CASE与HAVING一起使用，如下所示

DECLARE @test AS TABLE(EMPID INT, Badge VARCHAR(50), idNumber INT, EName VARCHAR(50))
INSERT INTO @test VALUES
(1148, '6542-74488', 66448, 'Adam Jhon'),
(1148, '642-8562', 66448, 'Adam Jhon'),
(1148, '3214-52874', 66448, 'Adam Jhon'),

(1149, '3214-45220', 209541, 'Tom Koyaski'),
(1150, '3214-23134', 63339, 'Shirin Abdulla'),
(1151, '3214-42355', 65498, 'Linda Jhon'),

(1151, '6542-2546', 65498, 'Linda Jhon'),
(1152, '3214-47632', 208673, 'Gayeth'),
(1153, '6542-73085', 83209, 'Maria Smith'),

(1153, '3214-58073', 65498, 'Maria Smith'),
(1154, '3214-26735', 208673, 'Ayan Jacob'),
(1155, '642-26739', 53959, 'Wo Li')

SELECT empid, Ename 
FROM @test 
WHERE badge LIKE '6542%' OR badge LIKE '3214%'
GROUP BY empid, Ename
HAVING COUNT (DISTINCT(CASE WHEN badge like '6542%' THEN 1 
                WHEN badge LIKE '3214%' THEN 2 END))>1

输出：

empid   Ename
1148    Adam Jhon
1151    Linda Jhon
1153    Maria Smith

Answer 5

只需这样做

    Select a.empid, a.name 
    from TableName as a
    inner join TableName as b on a.Empid = b.Empid and a.idnumber = b.idnumber and b.badge like '3214%'
    where a.badge like '6542%'