Need to know how to do this function.

[champbronc2]

$JoinDate = ("SELECT joindate FROM `users` WHERE `username` = CONVERT( _utf8 '$customer' USING latin1 ) COLLATE latin1_swedish_ci");

$checkpemaile = mysql_query("SELECT * FROM users WHERE referer='' AND visits>=3*DateDiff ("d", NOW(), $JoinDate)");

$pemail_existe = mysql_num_rows($checkpemaile);

 

OK, So the problem is bolded.

 

It keeps saying that something is wrong with that. I am trying to make it do this: Multiply 3 by the difference of days between Now and the date a member joined.

 

Can anyone correct the script, and explain?

[redarrow]

look at interval http://dev.mysql.com/doc/refman/5.0/en/date-and-time-functions.html#function_datediff

[champbronc2]

I did. I can't find any information on multiplication.

 

Any help?

[FIREBALL5]

$checkpemaile = mysql_query("SELECT * FROM users WHERE referer='' AND visits>=3*DateDiff ("d", NOW(), $JoinDate)");

 

It seems that you have several problems.

I think this is the corrected version:

$checkpemaile = mysql_query("SELECT * FROM users WHERE referer=\"\" AND visits>=" . 3*DateDiff ("d", NOW(), $JoinDate));

If DateDiff() is a MYSQL function, then I'm probably wrong. If it's a PHP function, then I might be right, except I don't know what parameters it requires.