thunder708

i have figured out what i have done wrong, i had changed your code to say foreach ($users as $u) { and changed the users file to say $users[0]["name"]='liam'; $users[0]["pass"]='lol'; but forgot to change the declaration to $users = array(); it was still $user = array(); my bad thank you anyway

$user = $_REQUEST['user']; // requests user from form $pass = $_REQUEST['pass']; // requests pass from form include("users.php"); foreach ($users as $u) { if ($u['name'] == $user && $u['pass'] == $pass) { echo "well don eyou have logged in "; } }

thanks, this works if i put the contents of users.php into the main file but doesnt work if i include the file... any ideas?

hi all! right i have two files, login.php and users.php. users.php <? $user = array(); // Set Users $user[0]["name"]='liam'; $user[0]["pass"]='lol'; $user[1]["name"]='mike'; $user[1]["pass"]='rofl'; ?> and in the login.php i have a form that you type in a user and pass and it logs you in, i know how to do this if i set the values within the login file itself but i want to be able to use this array i have made. so my question is, how do i search the array for the required user and pass when it is submitted from the form?

i got my else in the wrong place cant believe i didnt notice that lol <? session_start(); if(!session_is_registered(admin)){ $a=1;} function login($pass) { $pass = $_REQUEST['pass']; if($pass == 'mypass'){ session_start(); session_register(admin); header('location:media2.php'); echo "success!"; } else { $data = 'Wrong Password'; return $data; }} /* ?> <LINK href="style.css" rel="stylesheet" type="text/css"> <? */ $do = $_REQUEST['do']; $pass = $_REQUEST['pass']; if($a==1){ ?> Please Provide Your Password to gain Access<br> <form action="media2.php?do=login" method="post"> <input type="hidden" value="1" name="1"> <input type="password" name="pass"><input type="submit" value="login"> </form> <? if ($do == "login") { login($pass); } } else { if ($do == "cd") { cd(); } elseif ($do == "dvd") { cd(); } elseif ($do == "games") { cd(); } echo "<a href='?do=cd'>Cd</a> | <a href='?do=games'>Games</a> | <a href='?do=dvd'>DVD</a>"; } function cd() { echo "CD"; } function dvd() { $data = 'dvd'; return $data; } function games() { $data = 'games'; return $data; } thanks for all help though

i think i know where i have gone wrong....ill let you know once ive figured it out

sorry for delay i have been at college heres the code i have attempted and its not working and i know i have // before the stylesheet btw <? session_start(); if(!session_is_registered(admin)){ $a=1;} /* ?> <LINK href="style.css" rel="stylesheet" type="text/css"> <? */ if($a==1){ ?> Please Provide Your Password to gain Access<br> <form action="media.php?do=login" method="post"> <input type="hidden" value="1" name="1"> <input type="password" name="pass"><input type="submit" value="login"> </form> <? } else { $do = $_REQUEST['do']; $pass = $_REQUEST['pass']; if ($do == "login") { login($pass); } if ($do == "cd") { cd(); } elseif ($do == "dvd") { cd(); } elseif ($do == "games") { cd(); } echo "<a href='?do=cd'>Cd</a> | <a href='?do=games'>Games</a> | <a href='?do=dvd'>DVD</a>"; } function login($pass) { session_start(); if($pass == 'mypass'){ session_register('admin'); header('location:media.php'); } else { $data = 'Wrong Password'; return $data; } } function cd() { echo "CD"; } function dvd() { $data = 'dvd'; return $data; } function games() { $data = 'games'; return $data; } i even bypassed the login just to test out the other functions but when clicking on the link it doesnt show anything

i know how to do a login script, this script works if i refer it to a seperate page instead of using it as a function but i dont want that, i want it all on one page as i cant be doing with multiple pages for a simple application of php. i wouldnt be asking for help on a scripting forum if i could find it in a tutorial would i? if you can find me a suitable tutorial that uses form data and functions, be my guest.

sorry didnt want to make new thread cos its same code casuing the problem.... also, the password isnt being transferred to the login() function. the form action should be media.php?do=login, i have altered this in my code before anyone says anything any ideas?

never mind......found it lol Thank you for pointing it out tho =D

if its this bit $pass = $_POST[pass]; when i try $pass = $_POST['pass']; i get error : unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING

i keep getting the error: unexpected $end on line 56 which is the last line of my code, i have thoroughly checked otu the code makign sure all { having matching } but cant explain why i get this error. here is my code: <? session_start(); if(!session_is_registered(admin)){ $a=1;} ?> <LINK href="style.css" rel="stylesheet" type="text/css"> <? if($a==1){ ?> Please Provide Your Password to gain Access<br> <form action="media?do=login" method="post"> <input type="hidden" value="1" name="1"> <input type="password" name="pass"><input type="submit" value="login"> </form> <? } else { $do = $_REQUEST['do']; if ($do == "login") { login(); } elseif ($do == "cd") { cd(); } elseif ($do == "dvd") { cd(); } elseif ($do == "games") { cd(); } echo "<a href='?do=cd'>Cd</a> | <a href='?do=games'>Games</a> | <a href='?do=dvd'>DVD</a> | } function login() { $pass = $_POST[pass]; if($pass == 'mypass'){ session_register('admin'); header('location:media.php'); } else { $data = 'Wrong Password'; return $data; } } function cd() { $data = 'cd'; return $data; } function dvd() { $data = 'dvd'; return $data; } function games() { $data = 'games'; return $data; } any ideas?

just look at my last post ive edited it and tested it out on my own website and it works and btw either way that i have suggested works

Thank you for this help, i cant believe how stupid i was i missed out the ASC and DESC much thanks!

$_SERVER['SERVER_NAME'] = $_SERVER['SERVER_NAME']."/"; Try that my two previous attempts didnt work, if you saw it. it should but you could just have it as: $server = $_SERVER['SERVER_NAME']."/"; just to save typing time