Hi Again Gab Wrote my earlier response in the early hours whilst a bit tired so only spotted the first error in the query. No offense but there's actually so many errors I won't waste your or my time by listing them, apart from this one: You run the query as $numresults=mysql_query($query2); but the query is declared as $query = "select * from table where etc. This never works - ever! Trust me:-) Anyway - this is what I came up with: $name = trim($_GET['q']); $query = "SELECT * FROM student WHERE name LIKE '$name%' ORDER BY name "; On a table containing select * from student; +----+-----------+ | id | name | +----+-----------+ | 1 | mark | | 2 | jo | | 3 | steve | | 4 | John | | 5 | dave | | 6 | johnathon | | 7 | janet | | 8 | joanne | +----+-----------+ my query produces select * from student where name like 'jo%' order by name; +----+-----------+ | id | name | +----+-----------+ | 2 | jo | | 8 | joanne | | 4 | John | | 6 | johnathon | +----+-----------+ select * from student where name like 'john%' order by name; +----+-----------+ | id | name | +----+-----------+ | 4 | John | | 6 | johnathon | +----+-----------+ Really couldn't work out what you were trying to do with LIMIT - $s will always be empty the 1st time its tested so will always be allocated a value of 0, which once again will result in no results being returned. Also if you want to append it to the original $query use the concatenation operator .= Hope this has been helpful, but you really need to get hold of a good book on MySQL - I started with, & still use as a handy reference, the MySQL 5.0 Certification Study Guide published by MySQL AS. If you're a diligent student you could even become a MySQL 5.0 Certified Developer Good luck with the rest of your script Mark
