How to check namelist before signup

[Anon]

Employee names are already in the "checkit" table. A registration form's data will go to "signup" table on submit.

Need to add to first part of the registration script a routine that will check the registration form's submitted fname (firstname) and lname (last name) to find out
if it is on the "checkit" table list of employee names.

If submitted fname, lname are on checkit table list, registration process proceeds, else print error message, exit.

How do I do this? Some kind of select join statement, or what? I don't understand about the submitted form data -- where is it held, and how it is referred to in a join statement? The field data for fname and lname are not yet entered into the signup table, so how to make the select join statement?

Thanks for any suggestions -- Naporan

[preston_stone]

$query = mysql_query("SELECT * FROM checkit WHERE fname = '$fname' AND lname = '$lname'");

$rows = mysql_num_rows($query);

if ($rows > 0){

echo "Error - user already exists";

}

else {

echo "Success!";

}

// You can improve upon this, of course

[Anon]

Thank you for your help. I guess I didn't state my requirement very well. The query to the checkit table is meant to verify that the person registering with the form for access to protected pages is listed, first name and last name, as a company employee on the checkit table.

If the first name/lastname are on the checkit list, then the person can proceed with registration. If the first/last names are NOT on the checkit list, then process exits with error page.

Consider what this requirement is, I don't know how to use the

if ($rows < 0){

call or =0 or >0 -- specific enough. I need to be able to match the firstname/lastname from the form submission to the row on the checkit table list that contains the same first and last names.

I tried:

if (fname !== $fname) {
if (lname !== $lname) {
print "ERROR MESSAGE";
}
}
else {

}

(proceeds with registration process)

but apparently MySql doesn't know I'm asking it to check the form submission data for firstname/lastname against the checkit table rows of firstname/lastname.

MySql gives me the "not in the db list of names" error message each time, whether the form submitted firstname/lastname is in a checkit table row or not.

So, am stumped. Thanks again for your help, and if you have further suggestions, I am listening, that's for sure.

Thanks, Naparon

[Anon]

Had to go the long way round, still don't understand why, but this works:

$sql = "SELECT * FROM checkit WHERE fname = '$fname' AND lname = '$lname'";
$result = mysql_query($sql);

$myrow = mysql_fetch_array($result);

$first = $myrow["fname"];
$last = $myrow["lname"];

	if ($fname != $myrow["fname"])
	{

	if ($lname != $myrow["lname"])
	{
	echo "<center><br><br><br><table border=0 width=450><tr><td align=left><font size=2 face=arial font color=black>";
	echo "Your name cannot be found in the TB Woods employee database.";
	echo " Please go back to the signup form and type in your first and last name again, or";
	echo " contact your supervisor.</font></td></tr></table></center>";
	echo "<center><br><br><br><font size=2 face=arial font color=black><form>";
	echo "<input type=button value=\"<< Go Back\" onClick=\"history.go(-1)\"></form></font></center>";
	exit;
	}
		}

	elseif ($fname == $myrow["fname"])
	{
		if ($lname == $myrow["lname"])
		{


	}
		}

PROCEEDS FROM HERE TO GO THROUGH REST OF SCRIPT

Am certain there must be a slick way to do this with less bytes -- anyone? Would love to know.

Naporan