Condition in if() function failing to return false.

[NoUser2U]

Hi all,A week or 2 ago I was busy with a very basic register/login system, where userinfo is stored in a .txt file called users.txt. The login condition was: "If the username inserts firstname and lastname, and it matches some values in the .txt file, then login is success!". Below is the code:

<?php$users="C:/wamp/www/project_02_login2/files/users.txt";$fh=fopen($file, 'r');$lines=file($users, FILE_IGNORE_NEW_LINES);if(isset($_POST['firstname']) && isset($_POST['lastname'])){ // This does not work 100%	$firstname=$_POST['firstname'];	$lastname=$_POST['lastname'];		foreach($lines as $line){		list($ln, $fn)=explode(';', $line);		if($lastname==$ln && $firstname==$fn){ 			echo 'Login was successful!';			break;		} else { 			echo 'Login failed!';			break;		}		}} else {	echo 'Please enter your details below';}?>

When i go to the login-page for the first time when opening my site in the browser, then it does say 'Please enter your details below'. But when i test the login-page by submitting the form while keeping the firstname and lastname fields empty, then i get "Login failed" instead of 'Please enter your details below'.What is the cause of this?Thnx in advance!In case it is necessary to know, the .txt file contains:lastname1;firstname1lastname2;firstname2lastname3;firstname3lastname4;firstname4etc. etc.That's why i used list($ln, $fn)=explode(';', $line);

[[dx]]

Why not use SQL, or variables in PHP?$login_1 = "login-here";$pass_1 = "pass-here";$login_2 = "login2-here";$pass_2 = "pass2-here";etc.

[jeffman]

A database will not answer OP's question, and I think from other posts that OP is just now learning sql.It might be worth var_dump($_POST) just to see what is being sent to the server.

[thescientist]

perhaps use the empty function to test for the presence of firstname and lastname.

[[dx]]

A database will not answer OP's question, and I think from other posts that OP is just now learning sql.It might be worth var_dump($_POST) just to see what is being sent to the server.

For sure.. First edit script to work with SQL.

[NoUser2U]

A database will not answer OP's question, and I think from other posts that OP is just now learning sql.It might be worth var_dump($_POST) just to see what is being sent to the server.

Thank you Deirdre!I used var_dump on $_POST and it returned:

array(3) { ["firstname"]=> string(0) "" ["lastname"]=> string(0) "" ["submit"]=> string(7) "Submit!" }

So obviously the $_POST's for firstname and lastname were 0 characters long.So i edited the code slightly and inserted another if() function, so it looks like this now:

if(isset($_POST['submit'])){ // This now works!!	$firstname=$_POST['firstname'];	$lastname=$_POST['lastname'];		if(strlen($firstname) == 0 || strlen($lastname) == 0){		echo 'Please enter your details below';	} else {// REST OF THE CODE AS DESCRIBED IN THE FIRSTPOST}

Thank you for hinting the var_dump() function Deirdré's Dad!. I will now definately try to think about that function more when i encouter problems!

[jeffman]

An alternative to the isset function is the empty function. Like isset, it does not throw a warning if you pass it an unset variable. In your case it would tell you that your variables have been set but have no content. Something like:if (!empty($_POST['firstname']) && !empty($_POST['lastname']) ) {// check the login}else {// handle the mistake}

[iwato]

Thank you Deirdre! Thank you for hinting the var_dump() function Deirdré's Dad!. I will now definately try to think about that function more when i encouter problems!

Yeah, he taught me the same many moons ago, and I have since been able to answer many questions on my own. A very handy function and helpful person, indeed.Roddy

[jeffman]

Shucks, guys. Anyway, I'm pretty sure I learned how useful that function is from justsomeguy (no surprise there).