如何使用htmlunit获取修改后的ajax内容

Question

我想与您分享如何检索由ajax更改的html页面的内容。

以下代码返回旧页面。

public class Test { 

public static void main(String[] args) throws FailingHttpStatusCodeException, MalformedURLException, IOException, InterruptedException { 
    String url = "valid html page"; 
    WebClient client = new WebClient(BrowserVersion.FIREFOX_17); 
    client.getOptions().setJavaScriptEnabled(true); 
    client.getOptions().setRedirectEnabled(true); 
    client.getOptions().setThrowExceptionOnScriptError(true); 
    client.getOptions().setCssEnabled(true); 
    client.getOptions().setUseInsecureSSL(true); 
    client.getOptions().setThrowExceptionOnFailingStatusCode(false); 
      client.setAjaxController(new NicelyResynchronizingAjaxController()); 
    HtmlPage page = client.getPage(url); 
    System.out.println(page.getWebResponse().getContentAsString()); 
} 

}

这到底是怎么回事？

Answer 1

答案是page.getWebResponse（）赋予初始页面。

为了得到更新的内容，我们必须使用页面变量本身

package utils; 

import java.io.IOException; 
import java.net.MalformedURLException; 

import com.gargoylesoftware.htmlunit.BrowserVersion; 
import com.gargoylesoftware.htmlunit.FailingHttpStatusCodeException; 
import com.gargoylesoftware.htmlunit.NicelyResynchronizingAjaxController; 
import com.gargoylesoftware.htmlunit.WebClient; 
import com.gargoylesoftware.htmlunit.html.HtmlPage; 

public class Test { 

public static void main(String[] args) throws FailingHttpStatusCodeException, MalformedURLException, IOException, InterruptedException { 
    String url = "valid html page"; 
    WebClient client = new WebClient(BrowserVersion.FIREFOX_17); 
    client.getOptions().setJavaScriptEnabled(true); 
    client.getOptions().setRedirectEnabled(true); 
    client.getOptions().setThrowExceptionOnScriptError(true); 
    client.getOptions().setCssEnabled(true); 
    client.getOptions().setUseInsecureSSL(true); 
    client.getOptions().setThrowExceptionOnFailingStatusCode(false); 
    client.setAjaxController(new NicelyResynchronizingAjaxController()); 
    HtmlPage page = client.getPage(url); 
    System.out.println(page.asXml()); 
    System.out.println(page.getWebResponse().getContentAsString()); 
} 

} 

我发现暗示在以下链接

http://htmlunit.10904.n7.nabble.com/Not-expected-result-code-from-htmlunit-td28275.html

艾哈迈德Ashour雅虎你写的： 嗨，你不应该使用WebResponse，这意味着从 服务器获得实际内容。你应该使用htmlPage .asText（）或.asXml（）您的，Ahmed