MySql help

[maestro]

I want to join select 2 fields from 2 different tables, and the usual error that one of the fields is ambigous keeps coming up...I have used JOIN, no dice...any help will do. this is my mysql version *MySQL client version: 5.0.37*

[omonija]

shouldn't this be in the techie forum?

[bored]

C&P your query.

[bored]

it sounds like one of the fields is present in both tables. Did you specify which field is from which table as per table1.field1 and table2.field2 ?

[Arokoyo]

you have the precede the fieldname with the name of the table.
tablename.fieldname
it's ambigious cos the field probably exists in both tables, so its not sure which table you are referencing. make sure that all your fields are done that way!

[dibia]

...what bored said. The ambiguous error usually means you need to specify which table the troublesome column is in.

[Arokoyo]

ok didn't see that bored

[dibia]

Originally Posted by omonija

shouldn't this be in the techie forum?

...oh yeah, i almost forgot....@this post.

[virgoluv]

hmm geektalk...that was almost hot pipuz.

[djmightymike]

Classic relational DB problem.
consider two tables


and lets say you'd like to return lastnames of employees for which thier department id exists in the department table.

SELECT Employee.LastName
FROM Employee
JOIN Department
ON Employee.DepartmentID = Department.DepartmentID

pm me the table structures and the set you'd like to return if u still getting errors

[maestro]

This is my problem, i have named my fields table.column name, but i am still have some SQL errors.
What i actually want to achieve is allowing 2 people from 2 usergroups to be able to log into a page i.e peeps from an admin table and peeps from a superadmin table

this is my sql

"SELECT admin.Username, admin.Password, superadmin.Username, superadmin.Password 
FROM Admin 
LEFT JOIN superadmin 
USING (admin_id) 
WHERE admin.Username = '%s' AND admin.Password= '%s'"

:frustated

[djmightymike]

The first line in the query will fail becasue you're selecting superadmin columns from Admin table.

secondly, i think this problem can be easily solved with better db design. consider having a "privilege" key that will exist in both tables and then that way you can select a set using that "privilege" column as key.

brb..let me go complie this build i'm doin and i'll try fiind a solution for what u wanna do. My queries are based on SQL server but most likely it wil pass on mySQL

[maestro]

Originally Posted by djmightymike

The first line in the query will fail becasue you're selecting superadmin columns from Admin table.

secondly, i think this problem can be easily solved with better db design. consider having a "privilege" key that will exist in both tables and then that way you can select a set using that "privilege" column as key.

brb..let me go complie this build i'm doin and i'll try fiind a solution for what u wanna do. My queries are based on SQL server but most likely it wil pass on mySQL

...thank U..I just saw that error. Make I fix am letter.

[Arokoyo]

Originally Posted by maestro

This is my problem, i have named my fields table.column name, but i am still have some SQL errors.
What i actually want to achieve is allowing 2 people from 2 usergroups to be able to log into a page i.e peeps from an admin table and peeps from a superadmin table

this is my sql

"SELECT admin.Username, admin.Password, superadmin.Username, superadmin.Password 
FROM Admin 
LEFT JOIN superadmin 
USING (admin_id) 
WHERE admin.Username = '%s' AND admin.Password= '%s'"

:frustated

1. why are you using a left join?
2. you want to select admin.username and admin.password ending with "s". you use the keyword "like" not "=".

PHP Code:

SELECT Admin.Username, Admin.Password, SuperAdmin.Username, SuperAdmin.Password
FROM Admin INNER JOIN SuperAdmin ON Admin.adminId = SuperAdmin.adminId
WHERE (((Admin.Username) Like "%s")) OR (((Admin.Password) Like "%s")); 


or

PHP Code:

SELECT Admin.Username, Admin.Password, SuperAdmin.Username, SuperAdmin.Password
FROM SuperAdmin
LEFT JOIN Admin ON SuperAdmin.adminId = Admin.adminId
WHERE (((Admin.Username) Like "%s")) OR (((Admin.Password) Like "%s")); 


[Arokoyo]

Originally Posted by maestro

...thank U..I just saw that error. Make I fix am letter.

okor mi, tell me adminid is not common in both tables. na wa oh
anyways, im thinking of you
stalk you later