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Editorial Team
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Asked: 2026-06-14T15:12:09+00:00

I look up the image associated with a pokemon and display it with php.

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I look up the image associated with a pokemon and display it with php. Then I want to be able to “flip the card over,” by clicking on it. I’ve got the first click down, but the second click to flip the card back over isn’t working. I figure it’s the syntax of my php variable within the JS:

<!DOCTYPE HTML>
<html>
<head>
<meta charset="UTF-8">

<title>
'Murica!
</title>

<script>

function changeImage()
{

element=document.getElementById('pokemon_card')

if 
(element.src.match("http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-     back.jpg?w=750"))
{element.src="'.$result['image_url'].'";} //<- no idea how to express the php string variable here

else
{element.src="http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-back.jpg?     w=750";}

}

</script>

</head>

<body>

<?php

$dbhost = 'databasePlace';
    $dbname = 'mine';
    $dbuser = 'me';
    $dbpass = '******';

    $link = mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);

    mysqli_select_db($link,$dbname);    


$name = $_GET["fname"];



                $query = sprintf("SELECT image_url, Type
                                  FROM Pokemon c
                                  WHERE c.name='%s'",
                mysqli_real_escape_string($link,$name));

    $result = mysqli_fetch_assoc(mysqli_query($link,$query));

    echo '<img id="pokemon_card" onclick="changeImage()" height="225" 
width="165" src="'.$result['image_url'].'"/>';

mysqli_close($link);

?>

</body>
</html>
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1 Answer

  1. Editorial Team

    The easiest way would be to put a little PHP script inside the Javascript, like this…

    <script type="text/javascript">
function bla() {
    var thevar = "<?php echo $thevar; ?>";
}
</script>

    In other words, according to your question, you would replace the line {element.src="'.$result['image_url'].'";} with the line {element.src="<?php echo $result['image_url']; ?>";}

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