This Query Dose NOT work in Access 2000

SELECT (Members.First_Name + " " + Members.Last_Name)AS Member,
(SELECT Friend_E_Mail,
FROM    Friends,Members WHERE My_E_Mail = ? and Friend_E_Mail <> ?
UNION ALL
SELECT My_E_Mail,FROM  Friends,Members 
WHERE  Friend_E_Mail = ?and My_E_Mail <> ?) AS  E_Mail ,
(Members.First_Name) AS Name

FROM Members,Friends

Sample Tables

Members(all VARCHAR)     SOME DATA
First_Name               Alester Jude Carl Jones
Last_Name                A       B    C    J

FRIENDS(ALL VARCHAR)
My_E_Mail               Alester@lam.com Alester@lam.com  Alester@lam.com
Friend_E_Mail           jude@lam.com    carl@lam.com   jones@lam.com

Desired Output if ("?" in above query is: jones@lam.com)

+--------------+-----------+------------+
|Member        |E_Mail     |  Name      |
+---------------------------------------+
 Alester A   Alester@lam.com  Alester

Desired Output if("?" in above query is: Alester@lam.com)

+--------------+-----------+------------+
|Member        |E_Mail     |  Name      |
+---------------------------------------+
 Jude B        jude@lam.com  Jude
 carl C        carl@lam.com  Carl
 Jones J       jones@lam.com Jones

PS the "?" are query string parameters that im passing in the "?" i know that works fine.

MY QUESTION IS : i keep getting this error "Operation Not Allowed in Sub Query"

is their a work around query i can use without using a stored procedure or using multiple queries since this needs to be ONE SINGLE QUERY!?

Thanks.

有帮助吗?

解决方案

You can´t return more then 1 row in a select subquery. To this case you have to use the CASE expression.

You have to do something like this:

SELECT (m.First_Name + " " + m.Last_Name) AS Member,
       case 
       when f1.my_email is null 
        then f2.my_e_mail
        else f1.friend_email 
       end as email,
       m.First_Name AS Name
from members m
left outer join friends f1
  on m.email = f1.my_e_mail
  and f1.friend_e_mail = ?
left outer join friends f2
  on m.email = f2.friend_e_mail
  and f2.my_e_mail = ?;

Thys way it will return a row for each friend. You can see that the names of the fields are not so good, try to refactory it. my_e_mais isn´t so representative.

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