PHP & MySQL HELP!

[bethaliz]

Hi,

 

I've relatively new to the world of programming. I just started taking a course last year. I've learned a lot, unfortunately neither PHP or MySQL were taught to us. Now, we're in our 4th term and starting on project work. Our client wants a PHP & MySQL inventory management system, so we're learning on the fly.

 

MySQL, so far, hasn't turned out to be that difficult. It's mainly the same as SQL server, a few slight differences, but nothing worth talking about. We have our database, up and running, tested, all's well.

 

However, We're now starting to construct our PHP pages. My goal for the weekend, was simply to establish a connection to the database and run a query that returns a single value and output that value in a text box. Not exactly rocket science, but lets start easy.

 

Anyway. This is my code so far

 

<?php

 

$db_host = "localhost";

$db_user = "root";

$db_pass = "b123dw9WHT3";

$db_name = "company";

 

$db_cxnError = 'The connection to the database has failed: Please contact your System Administrator';

 

$cxn = mysql_connect($db_host, $db_user, $db_pass, $db_name)

or die($db_cxnError);

 

$query = "SELECT coName FROM contact WHERE coID=1";

$result_contact = mysql_connect($cxn, $query)

or die($db_cxnError);

 

?>

 

 

 

I have more in the HTML section, basically a textbox with the results of my query being echo'ed out as the value of the textbox.

 

When I try to run the page this is what I get

 

Warning: mysql_connect() expects parameter 1 to be string, resource given in C:\xampp\htdocs\signal\database_test.php on line 14

The connection to the database has failed: Please contact your System Administrator

 

parameter 1 is a string! So I'm not sure why it wont connect.

Any help you can provide would be great.

 

Also, I wasn't sure if this belongs here or in the SQL section, so feel free to move it if need be.

 

Thanks!

 

[phpretard]

<?php

$con = mysql_connect("localhost","root","b123dw9WHT3");

if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  
mysql_select_db("company", $con);

$result = mysql_query("SELECT coName FROM contact WHERE coID='1' ");

while($row = mysql_fetch_array($result))
  {
  $info=$row['coName'];
  
  echo "<input tpe="text" value="$info" />
  }
  
mysql_close($con);
?>