PHP and MySQL

[frog]

I am trying to test to see if a record exists before creating a new one. The problem is that even if the record does not exist my code is reporting that it does exist. I am very much a newbie and I am not sure if I can use !mysql_query in this way. Here is the code:

//connect to mysql database resourceplanner
$link_id = db_connect();
if(!$link_id) error_message(sql_error());

//test to see if the project number already exists

$query = "SELECT project_number FROM projects
WHERE project_number = $Project_Number";

if (!mysql_query($query))

{
//create a new projects record
$query = "INSERT INTO projects VALUES(NULL,
'$Project_Number',
'$Project_Description',
'$Producer')";

			if(!mysql_query($query))
			{
						error_message(sql_error());
			}
			else
			{
			 	message();
			}

}
else
{
error_message("Project Number $Project_Number already exists");
}

[cahva]

Use:

..
..
$query = "SELECT project_number FROM projects WHERE project_number = $Project_Number";
$res=mysql_query($query);
$numrows=mysql_num_rows($res); // Insert this to get the number of rows returned

if (!numrows)

{
//create a new projects record
$query = "INSERT INTO projects VALUES(NULL,
'$Project_Number',
'$Project_Description',
'$Producer')";
..
..

Or:

..
..
$query = "SELECT project_number FROM projects WHERE project_number = $Project_Number";
$res=mysql_query($query);

if (!mysql_result($res,0))

{
..
..

[frog]

Many thanks, the second code snippet works. I am going home now for a lie down