BlameThePixel!

I'm trying to learn how to use MySQL in PHP, but of course i have some problems. I've typed this in a file called mysql.php (minus the spaces) :

< html >
< body bgcolor="f6fff6" >
< font face="Verdana" style="font-size: 12px; font-color: #11ff11" >
< ?php
$con = mysql_connect("localhost", "root", "" ) ;
$db = mysql_select_db("test" ) ;

if(isset($_POST['name']) && isset($_POST['favcolor'] ) )
{
if(isset($_POST['name'])) $name_ = $_POST['name'];
if(isset($_POST['favcolor'])) $favcolor_ = $_POST['favcolor'];

$sql = "SELECT * FROM test WHERE name = ".$_POST['name'];
$result = mysql_query("$sql" ) ;

if(!($row = mysql_fetch_array($result))) {
$sql = "INSERT INTO test (name, favcolor) VALUES ('".$name_."', '".$favcolor_."' ) ";
$result = mysql_query("$sql" ) ;
}
}

echo "< table bgcolor='eeffee' width='400'

border='1' >< tr >< td >< b >Name< /b >< /td >< td >< b >Favorite color< /b >< /td >< /tr >";

$sql = "SELECT * FROM test ORDER BY name";
$result = mysql_query("$sql" ) ;

while($row = mysql_fetch_array($result)) {
echo "< tr >< td >".$row['name']."< /td >< td >".$row['favcolor']."< /td >< /tr >";
}

mysql_close($con ) ;
?>
< /table >
< br >< br >< br >< br >
< form method="POST" action="mysql.php" >
Name:             

< input type="text" name="name" >< br >
Favorite color:  < input type="text" name="favcolor" >< br >
< input type="submit" value="Submit" >
< /form >
< /font >
< /body >
< /html >

I'm trying to make a page where you can enter your name and favorite color in two boxes, and all names and colors typed in will appear in a table. But I want all names to be unique. But when I run the file, there is an error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in c:\program\php\easyphp\www\mysql.php on line 16. The database is called test, and the only table I use here is also called test. What is wrong, can anyone help?

you don't have to put the variables outside of quotes. Most of the time it's a lot less confusing if you leave them in...

anyway, change

\n\n

to

\n\n

I think that should work, not sure you can use backslashed "s in mysql but I don't see why you wouldn't be able to.
The variable has to be in quotes in the mysql, cause it's a string.

basically u need quotes around the name field for your query. You can tell zog doesn't use superglobals too much, cos the code he gave will give a T_variable error, you'll need the following:

\n\n

yeh your right, I rarely use them, which was one of my uncertainties.

Nope, doesn't work. It still says there is an error at line 16, the first mysql_fetch_array().

Edit: What tecnique do you use for BTP, ZoGgEr?

put echo mysql_error(); underneath the query line. If this prints nothing, the query is ok. If the error still occurs, try changing this:
\n\n
to this:
\n\n
Sorry if I got it wrong.

No, didn't work... It says there is an error with mysql_num_rows() instead. *sigh*

two approaches, try the query int phpmyadmin to see what it returns, or use my way of looping through the results from a select statement and for the first query since you're only checking for the existance of a row, the mysql count() function will do nicely:
\n

Err... Sorry, but that's too advanced for me, I don't understand. Isn't there an easier way?

erm, whats advanced about that??/

Didn't I say that I'm a n00b? Anyway, I've got an answer in a PHP forum which works.