Using $_POST in inline CSS

Question

As the title says, I am trying to use a PHP variable in inline CSS code. Here is what I have for the CSS:

<style>
    p {
    text-align: center;
    }

    img{ 
        -moz-animation:<?php $_POST["rmp"]; ?>s rotateRight infinite linear; 
        -webkit-animation:.6s rotateRight infinite linear; 
    }

    @-moz-keyframes rotateRight{
        0%{ -moz-transform:rotate(0deg); -moz-transform-origin:50% 50%; }
        100%{ -moz-transform:rotate(360deg); }
    }

    @-webkit-keyframes rotateRight{
        0%{ -webkit-transform:rotate(0deg); -webkit-transform-origin:50% 50%; }
        100%{ -webkit-transform:rotate(360deg); }
    }
</style>

Here is the HTML/PHP:

<body>
    <img src="http://upload.wikimedia.org/wikipedia/commons/thumb/3/38/Blue_circle_with_white_highlight.svg/547px-Blue_circle_with_white_highlight.svg.png" alt="wheel">

    <form action="wheel.php" method="POST">
        RPM: <input type="number" name="rmp">
             <input type="submit" value="Submit">
    </form>
    <?php
        echo $_POST["rmp"];
    ?>
</body>

DEMO (Works differently in Firefox and Chrome)

Hauke Ingmar Schmidt · Accepted Answer · 2014-01-09 23:31:51Z

3

You have to use print or echo, so change this:

<?php $_POST["rmp"]; ?>

To this:

<?php echo $_POST["rmp"]; ?>

PHP documentation on echo

edited Jan 9, 2014 at 23:31

Hauke Ingmar Schmidt

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answered Jan 9, 2014 at 23:21

Nick Coons

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Comments

Maaz · Accepted Answer · 2014-01-09 23:24:58Z

1

You need to use echo to actually output the value of the variable.

Furthermore, using inspect element on chrome you can easily see that for your img css you've got the following showing

-moz-animation:<?php $_POST["rmp"];

Instead of it outputting the actual value.

So use <?php echo $_POST["rmp"]; ?>

answered Jan 9, 2014 at 23:24

Maaz

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