query problem

[squiblo]

I have never attempted mysql like this before...this is what the table looks like..

 

||  user_id ||  other_person_id  ||  friend  ||  foe ||

        2                    44                      1          0

      34                      2                        1          0

 

 

What i want to do is this.... say that the $_SESSION['user_id'] == 2

 

and $user_id = $SESSION['user_id'];

 

Without doing 2 queries how can I pick put all the people that are friends with $user_id?

[squiblo]

I guess it would be a little like this with a few changes...

 

mysql_query("SELECT * FROM user_id AND other_person_id WHERE user_id='$user_id' AND other_person_id='$user_id'");

[squiblo]

i came up with this, but im not sure...

 

SELECT * FROM user_id AND page_zebra WHERE user_id='$user_id' AND page_zebra='$user_id'

[squiblo]

forget that previous post, thats just be being stupid once again...i think i ment this...

 

mysql_query("SELECT * FROM table_name WHERE user_id='$user_id' OR page_zebra_id='$user_id' AND friend=1");

 

i have tried this but it only give me data from 1 row

[aebstract]

 

mysql_query("SELECT * FROM table WHERE user_id = $user_id AND friend = '1'");

[squiblo]

Thank, but i also what to select all from where other_person_id = '$user_id'

[squiblo]

an easier way of putting it, how can i merge these 2 queries into 1...

 

mysql_query("SELECT * FROM table WHERE user_id='$user_id' AND friend=1");

 

and

 

mysql_query("SELECT * FROM table WHERE other_person_id='$user_id' AND friend=1");

 

 

and then my aim is to echo out the the row that doesnt have $user_id in it

 

 

NOTE: $user_id will never been the same in both rows

 

 

[fenway]

You can use an OR.