Code Issue

[TheHappyPeanut]

Without seeing what line 5 is (and the surrounding lines, really), I can't help you much.

Whoops

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<?php

var_dump($_POST)

if (!empty($_POST))
{
    if (isset($_POST['username']) && isset($_POST['password']))
    {
    $connect = mysql_connect("localhost","root","") or die("Could not connect");
    mysql_select_db("tracker") or die("Could not find database");
    }
    else
        die("Incorrect information");    
}
   
?>

Line 5 being

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if (!empty($_POST))

[Celauran]

You're missing the semicolon after var_dump

[TheHappyPeanut]

You're missing the semicolon after var_dump

Well, I'm dumb.

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array(2) { ["username"]=> string(0) "" ["password"]=> string(8) "password" }

[Celauran]

So $_POST['username'] is empty, which is why the if fails.

[TheHappyPeanut]

So what would that point to? An issue with my HTML markup or an issue with my database?

[TheHappyPeanut]

Actually, that's a flawed output because I didn't put anything in the log-in box (if I should have). Here's one where I filled in both fields.

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array(2) { ["username"]=> string(4) "John" ["password"]=> string(8) "password" }

[Celauran]

That one should work fine. You won't see any output to screen because you haven't done anything that produces any. You can echo some dummy text after selecting your database to confirm it's working if you like.

[TheHappyPeanut]

Like this?

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<?php


if (!empty($_POST))
{
    if (isset($_POST['username']) && isset($_POST['password']))
    {
    $connect = mysql_connect("localhost","root","") or die("Could not connect");
    mysql_select_db("tracker") or die("Could not find database");
    echo'testing';
    }
    else
        die("Incorrect information");
}
   
?>

[Celauran]

Sure. Using your previous form submission, you should see testing echoed.

[TheHappyPeanut]

Sure. Using your previous form submission, you should see testing echoed.

Yes, it's echoing the text when I attempt to log in.

[TheHappyPeanut]

I implemented mysqli a few minutes ago and I've not received any errors, so I'm assuming I've done it correctly. With that being said, the issue I'm having still exists.

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<?php

if (!empty($_POST))
{
    if (isset($_POST['username']) && isset($_POST['password']))
    {
    $mysqli = new mysqli();
    $mysqli = new mysqli('localhost', 'root', '', 'tracker');
    }
    else
        die();
}
   
?>

[Celauran]

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<?php

if (!empty($_POST))
{
    if (isset($_POST['username']) && isset($_POST['password']))
    {
        $mysqli = new mysqli('localhost', 'root', '', 'tracker');
    }
    else
        die();
}
   
?>

[Celauran]

With that being said, the issue I'm having still exists.

What issue is that? As per above, everything seems to be working fine given the little code you're written.

[TheHappyPeanut]

I can type anything I want into the form and it never gives me an error. Perhaps I've left a bit of code out that I require?

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if (isset($_POST['username']) && isset($_POST['password']))
    {
    $mysqli = new mysqli();
    $mysqli = new mysqli('localhost', 'root', '', 'tracker');
    }
    else
        die('Unable to proceed');

If I am correct, then this line of code basically says "If you don't provide me with a correct username or password, I'm closing the connection and telling you "Unable to proceed".

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else
        die('Unable to proceed');

[Celauran]

It's not performing any authentication. That block, in English, would ready "If a (any) username and password have been provided, open a connection to the database. Otherwise, don't and inform the user you couldn't." Or something to that effect.