FAQ: Introduction to PHP Form Validation - Validating Against Back-end Data

This community-built FAQ covers the “Validating Against Back-end Data” exercise from the lesson “Introduction to PHP Form Validation”.

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FAQs on the exercise Validating Against Back-end Data

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>   $username = $_POST["username"];
>   $password = $_POST["password"];
>   if(array_key_exists("${username}",$users)){
>     $passowrd == $user[$username] ? $feedback = $message : $feedback = $validation_error;
>   } else {$feedback = $validation_error;}
> };

tested my code, works perfectly fine.
yet, the error says ‘It looks like your code has a syntax error. Check that you’re not missing any semi-colons etc.’

Are you absolutely sure that your code is correct? When I try it, it lets me log in without a password.

1 Like

I am not saying what you do is wrong, but if we would take out the ternary operator we would have:

if (condition1)
   if (condition2) 

  } else {

} else {


where both else clauses do the same thing. why not go for:

if (condition1 and condition2)

} else {


Also, when using a ternary operator with assignment i would do:

$variable = condition ? true : false;

this eliminate the $feedback/$variable repetitiveness. Having gained all this insight, we could even do:

$feedback = condition1 && condition2 ? true : false;
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It didn’t come to my mind that this could be possible!
I changed the code to:

>   $username = $_POST["username"];
>   $password = $_POST["password"];
>    if(array_key_exists("${username}",$users)){
>     $passowrd === $user[$username] and $passowrd !== null ? $feedback = $message : $feedback = $validation_error;
>   } else {$feedback = $validation_error;}
> }

it fixes the bug, however, the error remains the same …

what value does and $passowrd !== null add? Seems reduntant

you have a typo in or more of your variable names. Codecademy lacks auto-complete/auto-suggest and spell check, so you need to be extra alert to that kind of mistake

The second step:

   $username = $_POST["username"];
   $password  = $_POST["password"];

What’s the purpose of a semicolon after the closing bracket?

No purpose. It’s unnecessary. It doesn’t do anything.

You can leave it out.

1 Like