Re: [PHP-DB] Hello

HI,
Thanks for your response. Here is my query. UserID is auto incrament  
and UserLastLogin is a current_timestamp.

$query_users = "INSERT INTO users(UserID, Username, UserEmail,  
UserPassword, UserFirstName, UserLastName, UserCompany, UserAddress,  
UserAddress2, UserCity, UserState, UserCountry, UserZip, UserPhone,  
UserFax, UserEmailVerified, UserRegistrationDate,  
UserVerificationCode, UserIP, UserLastLogin)
	
	VALUES('NULL','".$Username."','".$UserEmail."','". 
$UserPassword."','".$UserFirstName."','".$UserLastName."','". 
$UserCompany."','".$UserAddress."','".$UserAddress2."','". 
$UserCity."','".$UserState."','".$UserCountry."','".$UserZip."','". 
$UserPhone."','".$UserFax."','".$UserEmailVerified."','". 
$UserRegistrationDate."','".$UserVerificationCode."','".$UserIP."',  
now())";

This works as far as populating the database, but my results page  
does not return anything.

Only if the VALUES is set like this:

VALUES('NULL','".$Username=$_POST['Username']."','".$UserEmail=$_POST 
['UserEmail']."','".$UserPassword=$_POST['UserPassword']."','". 
$UserFirstName=$_POST['UserFirstName']."','".$UserLastName=$_POST 
['UserLastName']."','".$UserCompany=$_POST[$UserCompany]."','". 
$UserAddress=$_POST['UserAddress']."','".$UserAddress2=$_POST 
['UserAddress2']."','".$UserCity=$_POST['UserCity']."','".$UserState= 
$_POST['UserState']."','".$UserCountry=$_POST[$UserCountry]."','". 
$UserZip=$_POST['UserZip']."','".$UserPhone=$_POST['UserPhone']."','". 
$UserFax=$_POST[$UserFax]."','".$UserEmailVerified=$_POST 
[$UserEmailVerified]."','".$UserRegistrationDate=$_POST 
[$UserRegistrationDate]."','".$UserVerificationCode=$_POST 
['UserVerificationCode']."','".$UserIP=$_POST[$UserIP]."', now())";

but some do not work with this setup. variables like  
$UserEmailVerified, $UserRegistrationDate and $UserIP are not created  
from the form that was submitted.
for example, User IP date is created like this.

$UserIP = md5($_SERVER[REMOTE_ADDR]);

---------------------

Below is a snip of how I retrieve the info on the result page (dont  
want to clutter with whole code. also $fieldOne etc are MySql  
wildcards '%' from some dropdown lists that show before this code is  
executed. The results from adding show up fine there.)

$query_users = "SELECT * FROM users WHERE UserID LIKE '$fieldOne' AND  
Username LIKE '$fieldTwo' AND UserEmail LIKE '$fieldThree' AND  
UserPassword LIKE '$fieldFour' AND UserFirstName LIKE '$fieldFive'  
AND UserLastName LIKE '$fieldSix' AND UserCompany LIKE '$fieldSeven'  
AND UserAddress LIKE '$fieldEight' AND UserAddress2 LIKE '$fieldNine'  
AND UserCity LIKE '$fieldTen' AND UserState LIKE '$fieldEleven' AND  
UserCountry LIKE '$fieldTwelve' AND UserZip LIKE '$fieldThirteen' AND  
UserPhone LIKE '$fieldFourteen' AND UserFax LIKE '$fieldFifteen' AND  
UserEmailVerified LIKE '$fieldSixteen' AND UserRegistrationDate LIKE  
'$fieldSeventeen' AND UserVerificationCode LIKE '$fieldEighteen' AND  
UserIP LIKE '$fieldNineteen' AND UserLastLogin LIKE '$fieldTwenty'  
LIMIT $min, $max_results";

$result = mysql_query($query_users) or die(mysql_error());

for($i = 1; $i <= $num_sql; $i++) {
		$r = mysql_fetch_array($result, MYSQL_ASSOC);
		$UserID = $r['UserID'];
         	$Username = $r['Username'];
         	$UserEmail = $r['UserEmail'];
		$UserPassword = $r['UserPassword'];

so I have 3 pages. one adds the users, the next reviews and the last  
shows the results of what is picked.

Thanks,

Karl