Hide Form Div Using PHP if Statment

Question

Trying to make a login form become hidden once a successful login has happened within the form, so that the form cannot be seen by the user once they are logged in, but I still need a success message to be displayed when a successful login has happened. I have the form and the login sorted and it displayed a success message when a successful login has been made but I cant hide the form upon successful login. i have an if statement set up using isset but that's as far as I got I don't know what needs to go inside the if statement for me to hide the form.

This is the PHP code for my login page -

        <?php
    require('includes/application_top.php');
    $page_title='Login!';
    if (isset($_POST['action'])) {
        $username_input = $_POST['username'];
        $password_input = $_POST['password'];
        $login_ok = login($username_input, $password_input);
    }
    require('includes/site_header.php');
    ?>
    <style>
    <?php
    require('css/login.css');
    ?>
    </style>
    <br>
    <br>
    <br>
    <?php
            if (isset($login_ok)) {
                ?>
                <div class="row">
                    <div class="col-sm-12">
                        <?php
                        if ( ! $login_ok) {
                            ?>
                            <div class="alert alert-danger">
                                <p>Your credentials were invalid</p>
                            </div>
                            <?php
                        } else {
                            ?>
                            <div class="alert alert-success">
                                <p>Login successful</p>
                            </div>
                            <?php
                        }
                        ?>
                    </div>
                </div>
                <?php
            }
            ?>
    <!-- This is the if statement I mention -->
    <?php 
         if !isset($login_okay){
            ?>
            <div class="wrapper dafswInDown" hidden>
          }
    ?>


    <div class="wrapper fadeInDown">
      <div id="formContent">
        <!-- Icon -->
        <div class="fadeIn first">
          <br>
          <img src="images/Head1.jpg" style="height: 40%; width: 60%;" id="icon" alt="User Icon" />
          <br>
          <h3> Login! </h3>
        </div>
        <br>
        <!-- Login Form -->
        <form id="login_form" action="login_page.php" method="post">
          <input type="text" class="fadeIn second" id="username" name="username" placeholder="Username" value="<?= isset($_POST['username']) ? htmlspecialchars($_POST['username']) : '' ?>" maxlength="100" />
          <input type="password" class="form-control" id="password" name="password" placeholder="Password" value="" />
          <input type="submit" class="fadeIn fourth" name="action" value="Login"/>
        </form>

        <!-- Remind Passowrd -->
        <div id="formFooter">
          <a class="underlineHover" href="#">Forgot Password?</a>
        </div>


      </div>
    </div>


    <?php
    require('includes/application_bottom.php');
    require('includes/site_footer.php');
    ?>

@bromber it is probably wrong that was just as far as I had gotten before becoming stuck — Fawcett-Fawcett
– Fawcett-Fawcett, Commented Feb 23, 2021 at 13:37
@brombeer yes there should be brackets i have change this now — Fawcett-Fawcett
– Fawcett-Fawcett, Commented Feb 23, 2021 at 13:38
What doesn't work with your approach? Either use an opening <div> as you did (remember to close it after your form using the same if) or place your complete form inside the if so it doesn't get rendered at all. — user8034901
– user8034901, Commented Feb 23, 2021 at 13:38
the page doesn't load with what I have right now, would putting the entire div inside the if statement — Fawcett-Fawcett
– Fawcett-Fawcett, Commented Feb 23, 2021 at 13:39
Also take good care of naming stuff, you use $login_ok above but $login_okay in your if — user8034901
– user8034901, Commented Feb 23, 2021 at 13:40

return_false · Accepted Answer · 2021-02-23 13:50:42Z

1

There are a number of different ways you can approach this problem, but i think the best way to do this is with a session variable. Make your login function create a session variable that will follow the logged in user around throughout the website, then its just a matter of checking for the value and either displaying or hiding the form based on that. Here is a simplified example:

<?php function login($username_input=false, $password_input=false){
  // Do all your verification here
  // If logged in then:
  session_start();
  $_SESSION["user"] = true;
}

// Then on the form side just do this:
if(!$_SESSION["user"]){ ?> 
  FORM HERE
<?php } ?>

edited Feb 23, 2021 at 13:50

answered Feb 23, 2021 at 13:46

return_false

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12 Comments

Fawcett-Fawcett Over a year ago

How would i do it without the session?

return_false Over a year ago

You need the session. How are you going to know that the user is logged in when they move to another page?

Fawcett-Fawcett Over a year ago

as this is for a university project I will be coming to that part soon, is it a necessity to have a session?

return_false Over a year ago

Alternatively you can setup some kind of global variable but i wouldnt recommend it.

Fawcett-Fawcett Over a year ago

I attempted your way with the session however it didn't work, where should I have put things within the code as I put my form in where you said and it didn't work

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