Question

我正在尝试使用HTMLUnit使用Java浏览asp.net网站。当我在java代码的底部请求最终println时，控制台只返回部分代码而不是整个网页的html ...我已经尝试添加等待JavaScripts和其他常见的东西，如Xpath ...任何人都可以帮助？

这是我的代码：

public static HtmlPage submittingForm(String user, String pssw) throws Exception {              java.util.logging.Logger.getLogger("com.gargoylesoftware").setLevel(java.util.logging.Level.OFF);


    try (final WebClient webClient = new WebClient(BrowserVersion.CHROME)) {

        webClient.getOptions().setThrowExceptionOnScriptError(false);

        final HtmlPage page1 = webClient.getPage("someUrl");

        final HtmlForm form1 = page1.getFormByName("someForm");

        final HtmlTextInput username = form1.getInputByName("inputName");

        username.setValueAttribute(user);         

        final HtmlPasswordInput password = form1.getInputByName("someName");
        password.setValueAttribute(pssw);

        final HtmlPage page2 = (HtmlPage) form1.getInputByValue(" Login ").click();

        HtmlAnchor loginLink = page2.getAnchorByHref("anyHref");
        HtmlPage page3 = loginLink.click();

        final HtmlForm form2 = page3.getFormByName("formName");

        lastPage = (HtmlPage) form2.getInputByValue("buttonName").click();

        System.out.println(lastPage.asXml());

        }

    return lastPage;
}

Answer 1

根据我的经验，您可能只需要等待ajax完成加载网站所需的部分。

我使用这样的方法： //等待ajax不超过这个秒数 private static final int AJAX_MAX_TRIES_SECONDS = 30;

public static void waitForAjaxUntilTextAppearsInXmlSource(//
        @Nonnull final DomNode element, //
        @Nonnull final String text, //
        @Nonnull final String waitingLogMessage) throws WaitingForAjaxTimeoutException {
    LOGGER.debug("Waiting for ajax call to complete ... [" + waitingLogMessage + "]");
    final StringBuilder waitingdots = new StringBuilder("   ");
    for (int i = 0; i < AJAX_MAX_TRIES_SECONDS; i++) {
        if (element.asXml().contains(text)) {
            waitingdots.append(" ajax has finished [").append(text).append(" appeared in xml]");
            LOGGER.debug(" " + waitingdots);
            return;
        }
        waitingdots.append('.');
        wait(element);
    }
    LOGGER.warn(waitingdots.append(" ajax timeout [Text '" + text + "' did not appear in XML-source]").toString());
    LOGGER.warn("Page source:\n" + element.asXml());
    throw new WaitingForAjaxTimeoutException();
}

单击带有HTMLUnit的按钮只返回部分代码

1 个答案: