What is the Java equivalent of JavaScript's resource folder?

Question

My Wicket web application contains a Flash (*.swf) FLV player. The following code:

final String script = "var swfVersionStr = '10.0.0';"
    + "var xiSwfUrlStr = 'playerProductInstall.swf';"
    + "var flashvars = {file:'/proj/resources/video.flv'};"
    + "var params = {};"
    + "params.wmode = 'transparent';"
    + "params.quality = 'high';"
    + "params.allowscriptaccess = 'sameDomain';"
    + "params.allowfullscreen = 'true';"
    + "var attributes = {};"
    + "attributes.id = 'test';"
    + "attributes.name = 'test';"
    + "attributes.align = 'left';"
    + "swfobject.embedSWF('/proj/resources/mediaplayer.swf', 'movieDiv', '640', '480', swfVersionStr, xiSwfUrlStr, flashvars, params, attributes);"
    + "swfobject.createCSS('#flashContent', 'display:block;text-align:left;');";

add(new AbstractBehavior() {
    public void renderHead(IHeaderResponse response) {
        super.renderHead(response);
        response.renderOnLoadJavascript(script);
    }
});

plays the FLV. The swfobject.js file is placed in the resource folder of my server. As I am testing it on localhost, the absolute path of resource folder is: /home/tapas/Desktop/proj/work/Jetty_0_0_0_0_80_proj.war__proj__qk44r3/webapp . Now, how can I save a file in the resource folder of my server by using Java? JavaScript identifies the resource folder path as /proj/resources/ ; what is the equivalent expression of this path in Java? I have tried:

try{
    File file=new File("/proj/resources/joymaa.txt");
    if(file.exists()){
        System.out.println("File exists");
    }else{
        System.out.println("File does not exists");
    }
}catch(Exception exception){
    System.out.println(exception.getMessage());
}

This is not displaying any error message, but it shows "File does not exists."

Answer 1

I'd recommend not using an absolute file path to find a resource in a WAR file.

If your Java app needs a resource, best to put it in the CLASSPATH and use getResourceAsStream() from the servlet context to get an InputStream for reading.

Answer 2

try to find out what is your actual working directory of java... this might help: example I dont think that java could understand your relative path because it runs from different directory than javascript. Just my opinion, but give it a try.

Answer 3

（（（WebApplication）getApplication（））。getServletContext（）。getRealPath（“ / resources / video / xml / video.xml”）