CODEWITHSUNDEEP
phpmysql
Anna Garcia
Anna Garcia

Reputation: 37

php log in form does not respond

I'm trying what should be a pretty easy loginn form to a localhost database, but when I push log in it doesn't do anything. Basically I just get as input a username and password and try to get the data (I know the url looks weird as uid=password but it is as it has to be).

Then I compare what the database should return (name and uid) and compare those fields. Any hint would be of great help!

 <?php
        $log = file_get_contents("http://localhost:8080/users?uid=$password");
        $data = json_decode($log);
        $username= $_POST['username'];
        $password=$_POST['password'];
        
        if($log.empty) {
          header("Location: inici.html");
          exit;
        }
        
        if(($username == $data->Nombre) && ($password == $data->UID)){
          header("Location: inici.html");
          exit;
        }
        else
        {
          header("Location: login.html");
          exit;
        }
        ?>

Upvotes: 0

Views: 34

Answers (1)

Martin Zeitler
Martin Zeitler

Reputation: 76789

At first you write:

$data = json_decode($log);

and then you produce a run-time error with illegal/meaningless syntax:

if($log.empty) {

$log is a string, while you might have meant the $data array or object. In PHP there is only an -> object-access operator ... the . operator just concatenates.

You could check for: if(! empty($log)) to determine an empty response.

Upvotes: 2

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