SQL - 检查员工名字和/或姓氏是否存在

Question

根据下表，检查表 1 中的任一列是否存在于表 2 中 -

表_1：

FULL_NAME
Crystal, Crystal
Carmen, TEST2
XYZ, ABC
BLA, VVV

表_2：

NAME
Crystal
Carmen
ZZZ
AAA
VVV

输出：

FULL_NAME	STATUS
Crystal, Crystal	TRUE
Carmen, TEST2	TRUE
XYZ, ABC	FALSE
BLA, VVV	TRUE

Answer 1

您可以将exists 与instr 一起使用：

select t1.full_name, exists (select 1 from table_2 t2 where instr(t1.full_name, t2.name) > 0) status from table_1 t1;

输出：

full_name	status
Crystal, Crystal	1
Carmen, TEST2	1
XYZ, ABC	0
BLA, VVV	1

编辑：上面的解决方案将匹配任何出现的子字符串，但是，如果您希望匹配仅基于逗号分隔的值，您可以使用递归 cte 来获取子字符串：

with recursive vals(id, n) as (select row_number() over (order by (select 1)), t.* from table_1 t),
     cte(id, v, r) as (
         select id, case when instr(n, ", ") > 0 then substr(n, 1, instr(n, ", ")-1) else n end, case when instr(n, ", ") > 0 then substr(n, instr(n, ", ")+2, length(n) - instr(n, ", ")) else "" end from vals
         union all
         select id, case when instr(r, ", ") > 0 then substr(r, 1, instr(r, ", ")-1) else r end, case when instr(r, ", ") > 0 then substr(r, instr(r, ", ")+2, length(r) - instr(r, ", ")-1) else "" end from cte where length(r) > 0
)
select s1.n, s3.c from (select s.id, max(s.v in (select * from table_2)) c from cte s group by s.id) s3 join vals s1 on s3.id = s1.id;

见demo。

Answer 2

--Suported on MSSQL, MSACCESS, Oracle, MySQL, PostgreeSQL
SELECT t1.FULL_NAME,
CASE
    WHEN Temp.CONTAINING IS NULL THEN 'FALSE'
    WHEN Temp.CONTAINING IS NOT NULL THEN 'TRUE'
END AS 'STATUS'

FROM Table_1 t1
LEFT JOIN 
    (SELECT NAME, 
        (SELECT FULL_NAME
            FROM Table_1
            WHERE FULL_NAME LIKE CONCAT('%',NAME,'%')) as 'CONTAINING'
    FROM Table_2) Temp ON
t1.FULL_NAME = Temp.CONTAINING