我尝试访问基于发布数据返回后续页面的 ASPX 网站。不幸的是，我所有获取以下页面的尝试都失败了。希望这里有人知道在哪里可以找到错误！
在第一步中，我从 cookie 中读取会话 ID 以及返回的 html 页面中 viewstate 变量的值。第二步打算将其发送回服务器以获取所需的页面。
嗅探网络浏览器中的数据给出
Host=www.geocaching.com
User-Agent=Mozilla/5.0 (X11; U; Linux i686; en-US; rv:1.9.1.9) Gecko/20100618
Iceweasel/3.5.9 (like Firefox/3.5.9)
Accept=text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8
Accept-Language=en-us,en;q=0.5
Accept-Encoding=gzip,deflate
Accept-Charset=ISO-8859-1,utf-8;q=0.7,*;q=0.7
Keep-Alive=300
Connection=keep-alive
Referer=http://www.geocaching.com/seek/nearest.aspx?state_id=149
Cookie=Send2GPS=garmin; BMItemsPerPage=200; maprefreshlock=true; ASP.
NET_SessionId=c4jgygfvu1e4ft55dqjapj45
Content-Type=application/x-www-form-urlencoded
Content-Length=4099
POSTDATA=__EVENTTARGET=ctl00%24ContentBody%24pgrBottom%
24lbGoToPage_3&__EVENTARGUMENT=&__VIEWSTATE=%2FwEPD[...]2Xg%3D%
3D&language=on&logcount=on&gpx=on
目前，我的脚本看起来像这样
import java.net.*;
import java.io.*;
import java.util.*;
import java.security.*;
import java.net.*;
public class test1 {
public static void main(String args[]) {
// String loginWebsite="http://geocaching.com/login/default.aspx";
final String loginWebsite = "http://www.geocaching.com/seek/nearest.aspx?state_id=159";
final String POST_CONTENT_TYPE = "application/x-www-form-urlencoded";
// step 1: get session ID from cookie
String sessionId = "";
String viewstate = "";
try {
URL url = new URL(loginWebsite);
String key = "";
URLConnection urlConnection = url.openConnection();
if (urlConnection != null) {
for (int i = 1; (key = urlConnection.getHeaderFieldKey(i)) != null; i++) {
// get ASP.NET_SessionId from cookie
// System.out.println(urlConnection.getHeaderField(key));
if (key.equalsIgnoreCase("set-cookie")) {
sessionId = urlConnection.getHeaderField(key);
sessionId = sessionId.substring(0, sessionId.indexOf(";"));
}
}
BufferedReader in = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));
// get the viewstate parameter
String aLine;
while ((aLine = in.readLine()) != null) {
// System.out.println(aLine);
if (aLine.lastIndexOf("id=\"__VIEWSTATE\"") > 0) {
viewstate = aLine.substring(aLine.lastIndexOf("value=\"") + 7, aLine.lastIndexOf("\" "));
}
}
}
} catch (IOException e) {
e.printStackTrace();
}
System.out.println(sessionId);
System.out.println("\n");
System.out.println(viewstate);
System.out.println("\n");
// String goToPage="3";
// step2: post data to site
StringBuilder htmlResult = new StringBuilder();
try {
String encoded = "__EVENTTARGET=ctl00$ContentBody$pgrBottom$lbGoToPage_3" + "&" + "__EVENTARGUMENT=" + "&"
+ "__VIEWSTATE=" + viewstate;
URL url = new URL(loginWebsite);
URLConnection urlConnection = url.openConnection();
urlConnection = url.openConnection();
// Specifying that we intend to use this connection for input
urlConnection.setDoInput(true);
// Specifying that we intend to use this connection for output
urlConnection.setDoOutput(true);
// Specifying the content type of our post
urlConnection.setRequestProperty("Content-Type", POST_CONTENT_TYPE);
// urlConnection.setRequestMethod("POST");
urlConnection.setRequestProperty("Cookie", sessionId);
urlConnection.setRequestProperty("Content-Type", "text/html");
DataOutputStream out = new DataOutputStream(urlConnection.getOutputStream());
out.writeBytes(encoded);
out.flush();
out.close();
BufferedReader in = new BufferedReader(new InputStreamReader(urlConnection.getInputStream()));
String aLine;
while ((aLine = in.readLine()) != null) {
System.out.println(aLine);
}
} catch (MalformedURLException e) {
// Print out the exception that occurred
System.err.println("Invalid URL " + e.getMessage());
} catch (IOException e) {
// Print out the exception that occurred
System.err.println("Unable to execute " + e.getMessage());
}
}
}
知道有什么问题吗？非常感谢任何帮助！
更新
感谢您的快速回复！
我切换到使用 HttpURLConnection 而不是实现 setRequestMethod() 的 URLConnection。我还纠正了您提到的小错误，例如删除了过时的第一个 setRequestProperty 调用。
不幸的是，这并没有改变任何东西......我想我设置了所有相关参数，但仍然只获得列表的第一页。似乎“__EVENTTARGET=ctl00$ContentBody$pgrBottom$lbGoToPage_3”被忽略了。我不知道为什么。
在内部，网站上的表格如下所示：
<form name="aspnetForm" method="post" action="nearest.aspx?state_id=159" id="aspnetForm">
它由以下 javascript 调用：
<script type="text/javascript">
//<![CDATA[
var theForm = document.forms['aspnetForm'];
if (!theForm) {
theForm = document.aspnetForm;
}
function __doPostBack(eventTarget, eventArgument) {
if (!theForm.onsubmit || (theForm.onsubmit() != false)) {
theForm.__EVENTTARGET.value = eventTarget;
theForm.__EVENTARGUMENT.value = eventArgument;
theForm.submit();
}
}
//]]>
</script>
希望这有助于找到解决方案？
问候迈克。