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Home/ Questions/Q 8494949
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Editorial Team
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Editorial Team
Asked: June 10, 20262026-06-10T23:26:12+00:00 2026-06-10T23:26:12+00:00

I would like to display content only if the url contains certain parameter. In

  • 0

I would like to display content only if the url contains certain parameter. In the case is checked whether the URL exists, for example, the parameter ?offset=10. If the parameter exists (site.com/post-name?offset=10) is shown the content X, otherwise the content is shown Y.

Eg:

function parameterUrl() {
$str = "?offset=10";
$uri = $_SERVER['REQUEST_URI']; 
if ($uri == "http://site.com/post-name$str") {
echo "Show this";
}
 else {
}
} 

But the above function is not working. Can anyone help. Any idea is welcome. Thank you.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-10T23:26:13+00:00Added an answer on June 10, 2026 at 11:26 pm

    Your parameters are stored in the $_GET globals array.

    You need:

    if (isset($_GET['offset']) && $_GET['offset'] == 10)
    {
    echo "show this";
    }
    else {
    echo "show that";
    }
    

    Update from Comment

    If you are going to have multiple quantities then a switch statement would be better:

    if (isset($_GET['offset']))
    {
        switch($_GET['offset'])
        {
            case 10:
                echo "show for 10";
            break;
    
            case 20:
                echo "show for 20";
            break;
    
            case 30:
                echo "show for 30;
            break;
    //and so on
        }
    }
    else {
    echo "show for no offset";
    }
    
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