Dodgy Mysql and PHP Code :(

Anon

Hi everyone!
Can someone please check this code for me as it wont work and i have been trying for hours🙁

Thanx for your time,

Jonathan

================

<body><?php
$dbcnx = mysql_connect("localhost", "username", "password");
if (!$dbcnx) { echo( "<P>Unable to connect to the database server at this time. Please try again later.</P>" ); exit();}

mysql_select_db("catalog", $dbcnx);

if (! @mysql_select_db("catalog") ) { echo( "<P>Unable to locate database at this time.</P>" ); exit();}

$result = mysql_query("SELECT make, model, derivative, product, description, price FROM hsc_products WHERE
make LIKE '%$make%' AND
model LIKE '%$model%' AND
derivative LIKE '%$derivative%' and
description LIKE '%$description%'");

while($values_ar = mysql_fetch_array($result))
{
// enter the display text here
echo "$values_ar[make] ";
echo "$values_ar[model] ";
echo "$values_ar[derivative] ";
echo "$values_ar[description]";
}
?>

</head>

==============

[deleted]

NEVER use the @ sign.
In case of errors, print the mysql_error()
Don't quote variables.
Quote your strings.

echo "$values_ar[make] ";

should be

echo $values_ar['make'];

Anon

It dusnt bring up an error, just a blank screen.

[deleted]

Did you fix the errors I posted?

Anon

It doesn't give away any error messages because you are using the @.
That's why vincent told'ya not to use it.

What the @ does is ignore the error. Therefor no error is outputed.

Anon

I removed the @ and still no error is produced.

[deleted]

And did you fix the other things too??

Anon

Well, it could be an empty result set.

Try the following before the while loop:

if(sizeof(mysql_num_rows($result)) == 0) {
echo "0 results";
}