Question

its me with another trouble!

Table1

• n: 1
• n: 2
• n: 3

Table2

• n: 1
• n: 3

Query1

• numeber of results: 3

Query2

• numeber of results: 2

I need to compare this tables and if Table1 == Table2 echo “found”. So i made 2 diferent querys and Im doing it like this:

while($row1 = mysql_fetch_array($Query1))
{
while($row2 = mysql_fetch_array($Query2))
{
if($row1['n'] == $row2['n'] )
{
echo 'found';
}
}
}

Kinda dumb? :\ cause it seems show just the 1st result and stop.

Thanks

EDIT

Exmpl: I got this table: clients, and table: VIP clients. I need to search on table VIP clients if there is any client with same id, and result an echo: “found it”

Answer 1

Not sure if this is what you want, but you could do it in one query to see if there is any matching records in vipclients.

select a.*,b.* ,CASE WHEN b.clientid IS NOT NULL 
       THEN 'FOUND'
       ELSE 'NOT FOUND'
END AS vipexists 
from clients a left outer join vipclients b on a.clientid=b.clientid