AyKay47

MY EMAIL IS NUJASQUIRREL14@MSN.COM.. email me OP, one thing I suggest is that you check to make sure that all of the required fields are filled out before attempting to use them...something like. if(!empty($emailField = $_POST['email']) && !empty($emailField = $_POST['name']) && !empty($emailField = $_POST['company']) && !empty($emailField = $_POST['dept']) && !empty($emailField = $_POST['comment'])){ // continue to send mail }else{ // all required fields are not filled out.. } Also, what errors are you receving? do you have error_reporting set to E_ALL?

validate all form data with PHP, DO NOT use JS to validate any kind of user input as JS can be disabled, thus rendering your validation useless, thus allowing the user to enter anything that he/she desires into your form(s)

can you paste your code on this thread please..

instead of guess and check here, simply contact your host and they will guide you

sounds like a simple if else conditional statement is needed..checking if the user has chosen to upload a file or not

no homo...?

typically, whatever host you are using will provide the mail server for you

in your case, you would typically set the option value to the same text that is displayed to the user...most likely using a variable.. <select name="info"> <option value="<?php print $user; ?>"><?php print $user; ?></option></select>

then what is your purpose for this

you don't need to set up a mail server to let people send you messages through a form. All you need is and smtp server (which your ISP probably provides for you) and an email address. ...i thought he actually wanted to make one himself

1. as webstyles stated, your script will never work since you have $file substituted for the appropriate $_FILES array.. 2. why do you not want to upload a temp image, the image needs to be stored temporarily before it can be modified..

you want to spam... but you don't know how! ha!

http://www.google.com/#sclient=psy&hl=en&source=hp&q=php+mail+server+setup&aq=f&aqi=g1&aql=&oq=&pbx=1&bav=on.2,or.r_gc.r_pw.&fp=68b3f9ae28a95316&biw=1920&bih=999

well this will most likely be a unique login error, so anything that pertains to logging in/out

$ten_min_future = date('Y-m-d H:i:s', strtotime("+10 minutes"));

good catch..

what are the contents of $search coming from the URL?

i would probably do this with a db yes, set an if else condition checking if the user was successful in logging in or not...if they aren't successful...use an update query to update a field called something like "tries" and increment it by 1 each failed attempt

assuming that online_users and server_name are the columns that you need to extract values for..very basic here, class Core { public function server_status($col_name){ $query = mysql_query("SELECT * FROM server_status") // * being value $row = mysql_fetch_assoc($query,MYSQL_ASSOC); echo $row[$col_name]; } } $core = new Core(); $core -> $server_status("online_users"); $core -> $server_status("server_name");

yep

if you do not receive any mysql errors, i would try outputting your sql to see what its doing.. $qrytable1="SELECT images FROM images_list WHERE username=$username"; print $qrytable1;

do you mean that the query does not change if you enter another username? are your usernames actual strings like "username" or string numbers like "1"?

without seeing your code =

certainly can

1. why are you casting a username into an integer? is it a number string? 2. This will grab whatever images you have with the username of $username, seems that you are going about it correctly, I don't really understand the issue here and what you mean by "random images not showing"..?