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htmlunit: return a completely loaded page

Tags:

java

htmlunit

I am using HtmlUnit library for Java to manipulate websites programmatically. I can't find the working solution to my problem: How to determine that all AJAX calls are finished and return a completely loaded webpage? Here's what I have tried:

Firstly I create WebClient instance and make call to my method processWebPage(String url, WebClient webClient)

WebClient webClient = null;
    try {
        webClient = new WebClient(BrowserVersion.FIREFOX_3_6);
        webClient.setThrowExceptionOnScriptError(false);
        webClient.setThrowExceptionOnFailingStatusCode(false);
        webClient.setJavaScriptEnabled(true);
        webClient.setAjaxController(new NicelyResynchronizingAjaxController());
    } catch (Exception e) {
        System.out.println("Error");
    }
    HtmlPage currentPage = processWebPage("http://www.example.com", webClient);

And here is my method which should return a completely loaded web page:

private static HtmlPage processWebPage(String url, WebClient webClient) {
    HtmlPage page = null;
    try {
        page = webClient.getPage(url);
    } catch (Exception e) {
        System.out.println("Get page error");
    }
    int z = webClient.waitForBackgroundJavaScript(1000);
    int counter = 1000;
    while (z > 0) {
        counter += 1000;
        z = webClient.waitForBackgroundJavaScript(counter);
        if (z == 0) {
            break;
        }
        synchronized (page) {
            System.out.println("wait");
            try {
                page.wait(500);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }
    System.out.println(page.asXml());
    return page;
}

That z variable should return 0 if there are no JavaScript left to load.

Any thoughts? Thanks in advance.

EDIT: I found a partially working solution to my problem, but in this case I should know how the response page looks. For example, if a completely loaded page contains text "complete", my solution would be:

HtmlPage page = null;
    int PAGE_RETRY = 10;
    try {
        page = webClient.getPage("http://www.example.com");
    } catch (Exception e) {
        e.printStackTrace();
    }
    for (int i = 0; !page.asXml().contains("complete") && i < PAGE_RETRY; i++) {
        try {
            Thread.sleep(1000 * (i + 1));
            page = webClient.getPage("http://www.example.com");
        } catch (Exception e) {
            e.printStackTrace();
        }

    }

But what would be the solution if I don't know how a completely loaded page looks like?

like image 942
justasd Avatar asked Jun 06 '13 08:06

justasd


1 Answers

Try this:

HtmlPage page = null;
try {
    page = webClient.getPage(url);
} catch (Exception e) {
    System.out.println("Get page error");
}
JavaScriptJobManager manager = page.getEnclosingWindow().getJobManager();
while (manager.getJobCount() > 0) {
    Thread.sleep(1000);
}
System.out.println(page.asXml());
return page;
like image 91
brnfd Avatar answered Oct 19 '22 10:10

brnfd