PHP How To Show Hide DIVS based on Query String Value

Question

Essentially, I'm completely new at PHP - using dreamweaver cs6 (its all i have)

I have to get the value of a query string (someurl.com/id=something). After i get the value, I need to show a single div based on the ID.

While I know there really isn't any REAL "page load" event for PHP, I'd like to have it when the page opens.

So, how do I accomplish this?

I started with this:

<?php
    $stringId = $_GET['id']
   ?>

    <script language="javascript">      
        document.getElementById('$stringId').style.display = "block";
    </script>
    <?
?>

Since I don't have a server setup, will something like this work? Would you suggest something else? Am I completely wrong?

Any input would be appreciated. Thank you.

Answer 1

You should set the style inline with out relying on javascript as some users may have javascript turned off.

<div <?php if($_GET['id'] == 'X'){ ?>style="display: block;"<?php } ?>>
    your div content
</div>

with an else as requested

<div style="<?php if($_GET['id'] == 'X'){ ?>display: block;<?php } else { ?>display: none;<?php } ?">
    your div content
</div>

you could also use a predefined css class

based on your comment about having way to many to do a switch if you need to mix it up just insert this

<script language="javascript">      
    document.getElementById('<?php echo $_GET['id']; ?>').style.display = "block";
</script>

Answer 2

It should be something like this:

<?php
switch($_GET['id']) {
    case 'asdf':
        echo '<div>ASDF in QS</div>';
        break;
    case 'bcd':
        echo '<div class="some-class-because-i-can">something else</div>';
}

You can't simply mix javascript with php.

And of course, you need servers setup, server built in php(http://php.net/manual/en/features.commandline.webserver.php) should be enough for your needs.

Answer 3

PHP is server-side scripting language, meaning it only runs on the server (any output being rendered on the users screen), that being said you need a local server like WAMP XAMP to debug/ run your code.

What you are trying to do is not going to work PHP is executed on the server as i previously mentioned and Javascript runs in the viewers browser, the variable $stringID will not be present when your javascript code runs.

Why bother hiding an element that has the url ID when PHP can render that element and only that element.

<?php
    if(isset($_GET['id'])) {   //checks weather the link has data in the id field
       $stringId = $_GET['id'];

       echo "<div> content here </div>"; //anything you echo will be rendered in the 
                                         //pages source code be it HTML, Javascript etc.
    }
>?

If you want to fetch this data from php without refreshing the page use AJAX from jQuery (google it :D)