MadTechie

SO in the SQL statement use something like SELECT * FROM table1 ORDER BY field1, field2

well im going to ignore this Whole post as you haven't read the rules.. USE [ CODE] TAGS.

erm... English please also WTF change if(!\$_POST) { to if(!$_POST) {

without the code its going to be guess work so i suggest post the code as for the guess work, check the {} (mainly the last one)

Why do you say its ignoring the isset? the submit will always be set when the form is posted!

to output an array <?php print_r($sectors); ?> or <?php print_r($$_SESSION); ?>

Erm you mean " Order By " something else? can you show some sample code please

javascipt

humm this is kinda weird <?php session_start(); // Define your username and password $username = "username"; $password = "password"; $_SESSION['txtUsername'] = $_POST['txtUsername']; $_SESSION['txtPassword'] = $_POST['txtPassword']; if ($_SESSION['txtUsername'] != $username || $_SESSION['txtPassword'] != $password) { ?> if the username is correct it will login even if password is wrong! not sure how thats is meant to work!

echo some javascript to redirect

change <?php exec('unzip $filename',$ret); ?> to <?php exec("unzip $filename",$ret); ?> (Note ' to ") you can't parse a variable in single quotes

Yep try had it in a database try switching to 'utf8_general_ci' encoding

In unzip a valid program command on your server ? is exec allowed ? any errors ?

try <?php $theData = //<---The Date your viewing $query1= "SELECT COUNT(*) as Events FROM calendar WHERE event_start = $theData"; //get the "Events" number for the total number ?>

your have to use if on the display or setting it, ie <?php isAdmin = false; $details = array(IP => "11.22.33.44", Name => "MadTechie"); echo $details['Name'].""; echo (isAdmin? "IP:".$details['IP']:"" ?> UNTESTED

ok don't try my test..

so what errors are you getting ? if not try adding error_reporting(E_ALL);

What, do you mean by "the same" what was the result also where did user2 come form!

Simple test #2 <?php phpinfo(); ?>

what happens if you try this Create new user (note the static name and password INSERT INTO members_info (members_id,user_name,first_name,last_name,gender,birth_month,birth_day, birth_year,contact_number,email_address,register_date,password) VALUES(NULL,'test','$first_name','$last_name' ,'$gender','$birth_month','$birth_day','$birth_year','$contact_number','$email_address',Now(),MD5('test'))"); ?> then SELECT * FROM members_info WHERE user_name='test' AND password=MD5('test');

and the one is the database is ? also whats the field type in the database?

note change <?php $query = "INSERT INTO Person('FirstName', 'LastName', 'Age'); ?> to <?php $query = "INSERT INTO Person('FirstName', 'LastName', 'Age')"; ?>

True, @Glyde your script won't work unless you meant \s+ then again are we going by the title of the post or the contains LOL

i would do this, to keep the standard <?php $password = $_POST['password']; $result = mysql_query("select * from members_info where user_name='$user_name' and password =MD5('$password')"); ?> have you compared the md5 results?

another option <?php $string = "a. Testing <3>"; $patten = "<(.+)>"; ereg($patten, $string, $found); echo $found[1]; ?>