Learner, now I'm getting:
Quote:
Parse error: parse error, unexpected T_STRING, expecting ',' or ';' in c:\program files\easyphp\www\test.php on line 9
The code i'm using is:
Code:
<?php
$db = mysql_connect("localhost", "[user]", "[pass]");
	mysql_select_db("applications",$db);
		$result = mysql_query("SELECT * FROM applications");

		while($result2 = mysql_fetch_array($result)){

			echo "Name 
\n ".$result2['name'].";
			echo "Nick 
\n ".$result2['nick'].";
			echo "Telephone 
\n ".$result2['telephone'].";
			echo "Cellular 
\n ".$result2['cellular'].";
			echo "Date 
\n ".$result2['date'].";
			echo "Chance 
\n ".$result2['chance'].";
			echo "Complete 
\n ".$result2['complete'].";
			echo "Address 
\n ".$result2['address'].";
			echo "City, ST, Zip 
\n ".$result2['citystzip'].";
			echo "Age 
\n ".$result2['age'].";
			echo "E-Mail 
\n ".$result2['email'].";
			echo "AIM 
\n ".$result2['aim'].";
			echo "IRC 
\n ".$result2['irc'].";
			echo "UBB 
\n ".$result2['ubb'].";
			echo "Username 
\n ".$result2['username'].";
			echo "Password 
\n ".$result2['password'].";
			echo "Sections 
\n ".$result2['sections'].";
			echo "Birth Year 
\n ".$result2['yob'].";
			echo "Admin 
\n ".$result2['admin'].";
			echo "Staff 
\n ".$result2['staff'].";
			echo "Wants Admin 
\n ".$result2['wantadmin'].";
			echo "Wants Mod 
\n ".$result2['wantmod'].";
			echo "Wants News 
\n ".$result2['wantnews'].";
			echo "Was Admin 
\n ".$result2['wasadmin'].";
			echo "Was Mod 
\n ".$result2['wasmod'].";
			echo "Was News 
\n ".$result2['wasnews'].";
			echo "Notes 
\n ".$result2['notes'].";
		}
?>
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