基于Javascript的动态内容使用htmlUnit [英] Javascript based dynamic content using htmlUnit

问题描述

我一直坚持使用HtmlUnit获取基于JavaScript的动态内容。我期待从页面获得（Signin，注册html内容）。使用以下代码，我只获取静态内容。

I have been stuck in getting JavaScript based dynamic content using HtmlUnit. I am expecting to get (Signin, Registration html content) from the page. With the following code, I only get the static content.

我是HtmlUnit的新手。任何帮助都将受到高度赞赏。

I am new to HtmlUnit. Any help will be highly appreciated.

String strURL = "https://www.checkmytrip.com" ;
java.util.logging.Logger.getLogger("com.gargoylesoftware.htmlunit").setLevel(java.util.logging.Level.OFF);
java.util.logging.Logger.getLogger("org.apache.http").setLevel(java.util.logging.Level.OFF);

final WebClient webClient = new WebClient(BrowserVersion.FIREFOX_31);
webClient.getOptions().setJavaScriptEnabled(true);
webClient.getCookieManager().setCookiesEnabled(true);
webClient.waitForBackgroundJavaScript(60 * 1000);
webClient.setAjaxController(new NicelyResynchronizingAjaxController());

HtmlPage myPage = ((HtmlPage) webClient.getPage(strURL));

String theContent = myPage.getWebResponse().getContentAsString();
System.out.println(theContent);      

推荐答案

两点：

1. 获取页面后需要waitForBackgroundJavaScript（），如提示这里



2. 您应该使用myPage.asText（）或.asXml（）代替，因为getWebResponse（）返回原始内容而不执行JavaScript 。

1. You need to waitForBackgroundJavaScript() after you get the page, as hinted here

2. You should use myPage.asText() or .asXml() instead, because getWebResponse() returns the original content without JavaScript execution.

String strURL = "https://www.checkmytrip.com" ;
java.util.logging.Logger.getLogger("com.gargoylesoftware.htmlunit").setLevel(java.util.logging.Level.OFF);
java.util.logging.Logger.getLogger("org.apache.http").setLevel(java.util.logging.Level.OFF);

try (final WebClient webClient = new WebClient(BrowserVersion.FIREFOX_31)) {
    webClient.setAjaxController(new NicelyResynchronizingAjaxController());

    HtmlPage myPage = ((HtmlPage) webClient.getPage(strURL));
    webClient.waitForBackgroundJavaScript(10 * 1000);

    String theContent = myPage.asXml();
    System.out.println(theContent);
}

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