Mysql Query and PHP - Newbie - PHPBuilder Forums

Mysql Query and PHP - Newbie

Anon

Hi there!

I have produced the following code to extract data feom a sql database using a query which i believe to be correct;

The thing is i do not know what text to write next to display the results on a php page.

Please help im despirate!!

===========coding=================

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

mysql_select_db("catalog", $dbcnx);

if (! @mysql_select_db("catalog") ) { echo( "Unable to locate database at this time." ); 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%";");

=========== end ===============

thanx🙂

Jonathan

cordex

Okay, you've got the results, but they aren't in a displayable form yet. Now you've got to place them in an array.

while($values_ar = mysql_fetch_array($result))
{
// enter the display text here
echo "The value of field 'name' is $values_ar[name] ";
echo "The value of field 'address' is $values_ar[address] ";
// this will cycle through each record until the end is reached.
}

Is taht what you're looking for?

Anon

Your the man ben!

Looks like its workin but now it says:

"Error performing query: Query was empty".

The script now looks like:

=========begin=======

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

mysql_select_db("catalog", $dbcnx);

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

if (!$result) { echo("Error performing query: " . mysql_error() . ""); exit(); }

while($values_ar = mysql_fetch_array($result))
{
// enter the display text here
echo "The value of field 'name' is $values_ar[make] ";
echo "The value of field 'address' is $values_ar[model] ";
// this will cycle through each record until the end is reached.
}
?>

=========end=========
Thanx🙂

Jonathan

cordex

Looks like your query isn't getting any results from your database. That's a problem with the query itself, not the PHP code. In other words, it's not matching the query with any actual values in the DB.

blackmamba

Well there's a few mistakes in your query, try this instead:

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

I've replaced double quotes with singles (to prevent corrupting the string) and removed the semi-colon at the end of the string (only used in command line interface, not mysql_query() )