单击HtmlUnit按钮

时间:2013-09-01 07:23:53

标签: java button htmlunit

我正在尝试在www.meetme.com上发送消息,但无法弄清楚如何做到这一点。我可以在评论区域中输入消息,但单击“发送”按钮不会执行任何操作。我究竟做错了什么?当我登录并按下登录按钮时,页面确实会发生变化,一切都很好。有人有任何想法或线索吗?

    HtmlPage htmlPage = null;
    HtmlElement htmlElement;
    WebClient webClient = null;
    HtmlButton htmlButton;
    HtmlForm htmlForm;
    try{

        // Create and initialize WebClient object
        webClient = new WebClient(BrowserVersion.FIREFOX_17 );
        webClient.setCssEnabled(false);
        webClient.setJavaScriptEnabled(false);
        webClient.setThrowExceptionOnFailingStatusCode(false);
        webClient.setThrowExceptionOnScriptError(false);
        webClient.getOptions().setThrowExceptionOnScriptError(false);
        webClient.getOptions().setUseInsecureSSL(true);
        webClient.getCookieManager().setCookiesEnabled(true);

        /*webClient.setRefreshHandler(new RefreshHandler() {
            public void handleRefresh(Page page, URL url, int arg) throws IOException {
                 System.out.println("handleRefresh");
            }

        });*/

         htmlPage = webClient.getPage("http://www.meetme.com");
         htmlForm = htmlPage.getFirstByXPath("//form[@action='https://ssl.meetme.com/login']");

         htmlForm.getInputByName("username").setValueAttribute("blah@gmail.com");
         htmlForm.getInputByName("password").setValueAttribute("blah");

         //Signing in
         htmlButton = htmlForm.getElementById("login_form_submit");
         htmlPage = (HtmlPage) htmlButton.click();

         htmlPage = webClient.getPage("http://www.meetme.com/member/1234567890");

         System.out.println("BEFORE CLICK");
         System.out.println(htmlPage.asText());


         //type message in text area
         HtmlTextArea commentArea = (HtmlTextArea)htmlPage.getFirstByXPath("//textarea[@id='profileQMBody']");
         commentArea.setText("Testing");        


         htmlButton = (HtmlButton) htmlPage.getHtmlElementById("profileQMSend");
         htmlPage = (HtmlPage)htmlButton.click();
         webClient.waitForBackgroundJavaScript(7000);

         //The print is exactly the same as the BEFORE CLICK print
         System.out.println("AFTER CLICK");
         System.out.println(htmlPage.asText());

      }catch(ElementNotFoundException e){
        e.printStackTrace();
      }catch(Exception e){
        e.printStackTrace();
      }

1 个答案:

答案 0 :(得分:4)

在不了解您正在访问的网页的情况下,您无法在禁用JavaScript的情况下执行AJAX请求。如果更改不会导致成功,那么您将不得不继续调试,但要确保启用了JavaScript。

此外,请确保您使用的是HtmlUnit 1.12并更新代码中所有已弃用的方法。

顺便说一下,我还建议关闭JavaScript警告。检查this answer,了解如何做到这一点。