1) No. You must add third column that unically describes a person (primary
key). E.g. there will more than one john smith.
2) And my first answer is the answer on your first question:
table - fammem
personid famid memid
1 1 1
2 1 3
3 32 4
4 5 4
5 3 6
6 4 1
sql:
insert into fammem( famid , memid) values (5, 1);
insert into fammem( famid , memid) values (1, 1);
insert into fammem( famid , memid) values (3, 3);
...
the personid is autoincr. field.
3) You shouldn`t use INNER JOIN, you may use simple SELECT:
select fammem.*, family.name, members.name from family , fammem, members
where family.id = fammem.famid and fammem.memid=members.id
4) uffff....
"Neil" <[EMAIL PROTECTED]> �������/�������� � �������� ���������:
news:[EMAIL PROTECTED]
> I am no database guru and I am also new to php/mysql
>
> I want to be able to store data where one of the fields could have
multiple
> values which are stored in another table
>
> eg. the first table is called family with a field called name which
contains
> family names
> the second table is called members with a fiedl also called name which
> contains first names
> the third table is an intermediate table that only contains ids of the
other
> tables.
> (the data is below)
>
> By using inner joins I can get a list of all the people in the database
(or
> other variants eg. all the people with first names of Mary)
>
> My first question is - is this the best way to do stuff like this?
> My second question is - how do I insert new data? eg to insert Mary
Stevens
> (the first name Mary already exists)
>
> I hope this all makes some sense
>
> Thanks
>
> Neil
>
>
>
> table - family
> id name
> 1 jones
> 2 smith
> 3 mcdonald
> 4 talbot
> 5 jackson
>
>
> table - members
> id name
> 1 john
> 2 mary
> 3 ted
> 4 sally
> 5 bob
> 6 jane
>
>
> table - fammem
> famid memid
> 1 1
> 1 3
> 2 4
> 5 4
> 3 6
> 4 1
>
>
> SELECT * FROM family inner join fammem on family.id = fammem.famid inner
> join members on fammem.memid = members.id
> id name famid memid id name
> 1 jones 1 1 1 john
> 1 jones 1 3 3 ted
> 2 smith 2 4 4 sally
> 3 mcdonald 3 6 6 jane
> 4 talbot 4 1 1 john
> 5 jackson 5 4 4 sally
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