Is it possible to get URL parameters in JavaFX?

Question

You can get url parameters from the page: Get escaped URL parameter

function getURLParameter(name) {
    return decodeURIComponent((new RegExp('[?|&]' + name + '=' + '([^&;]+?)(&|#|;|$)').exec(location.search)||[,""])[1].replace(/\+/g, '%20'))||null;
}

Once you have the parameters you can pass the parameters to the embedded JavaFX application using the JavaFX deployment toolkits DTJava.js functions. See section 7.3.3 Pass Parameters to a Web Application of the JavaFX deployment guide.

<!-- Example 7-7 Pass Parameters to an Embedded Application -->

<head>
    <script type="text/javascript" src="http://java.com/js/dtjava.js"></script>
    <script>
        function deployIt() {
            // deployment guide sample modified by me to show 
            // getting the zipcode parameter from the url instead of hardcoding it.
            //var zipcode = 95054; 
            var zipcode = getURLParameter("zipcode");

            dtjava.embed(
                {            id: "myApp",
                            url: "Map.jnlp",
                          width: 300,
                         height: 200,
                    placeholder: "place",
                    params: {
                              mode: "streetview",
                              zip: zipcode
                    }
                },
                { javafx: "2.1+" },
                {}
            );
        }
        dtjava.addOnloadCallback(deployIt);
    </script>
</head>
<body>
    <div id="place"></div>
</body>

http://docs.oracle.com/javafx/2/deployment/deployment_toolkit.htm#BABJHEJA

As the JavaFX deployment guide says, to access parameters in the application code, use the getParameters() method of the Application class. For example:

String zipcode = app.getParameters().getNamed("zip");