tabithaboof

[!--quoteo(post=371459:date=May 5 2006, 01:08 AM:name=Prismatic)--][div class=\'quotetop\']QUOTE(Prismatic @ May 5 2006, 01:08 AM) [snapback]371459[/snapback][/div][div class=\'quotemain\'][!--quotec--] Close, but to get the variable from the form you must use $_POST, like the following: [code] $query = "SELECT * FROM tbl WHERE username='$_POST[user]'"; [/code] [/quote] Thank you SO much, you have no idea how long I have been fiddling with this with no luck.

I am quite new to PHP/MYsql and I have a site for high scores on a video game based around a simple MYsql database. something like this username, level1, level2, level3 bob , 1000 , 2000 , 3000 kate , 2000 , 3000 , 4000 I am trying to create a form to let users query the database to find out all the scores for a particular player. I am doing this using an HTML input form: <form method="post" action="scoreuser.php"> search form:<input type=text name='user' size=20 maxlength=20><input type=submit> </form> Based on what I have read "user" should be a variable I can use in the following script for the page "scoreuser.php" $query = "select * from tbl WHERE username='user'"; $result = mysql_db_query("scoreattack", $query); if ($result) { while ($r = mysql_fetch_array($result)) { $username = $r["username"]; $level1 = $r["level1"]; echo " $username $level1" Unfotunately this doesnt seem to work at all, if I enter "bob" into my form and submit the query I get a blank page. However if I change this line: $query = "select * from tbl WHERE username='user'"; to $query = "select * from tbl WHERE username='bob'"; Than I can display bobs scores. I am assuming I am not connecting the form to "scoreuser.php" correctly or I am getting the syntax for the 'user' variable wrong somewhere. I am sorry if this is a slightly long query but I wanted to be as clear as possible. Any help would be most greatly appreciated.