asp.net - http POST 与 asp.net

Question

我有一个烦人的 ASP.NET 问题:

我有一个 Perl 脚本(见下文)，它获取 form_info 变量。不幸的是，现在是 http POST，而不是 http GET，所以 Request.Querystring 不起作用...

现在我必须用 asp.net 页面/应用程序替换 Perl 脚本，但我的问题是当我没有字符串时我无法处理字符串 form_info...而且我无法将 http POST 更改为 HTTP get，因为它是由第 3 方 java 小程序生成的。

# Print out a content-type for HTTP/1.0 compatibility
print "Content-type: text/html\n\n";
#
#test whether it's via a firewall (i.e. GET multiple times)
# or direct, i.e. POST
$method = $ENV{'REQUEST_METHOD'};
if ($method eq "GET") {    
    $form_info = $ENV{'QUERY_STRING'};
 print LOGFILE "Method found was: REQUEST_METHOD\n";
}
elsif ($method eq "POST"){
    # Get the input
    $data_size = $ENV{'CONTENT_LENGTH'};
    read(STDIN,$form_info,$data_size);
 print LOGFILE "\nMethod found was: POST\n";
}
else {
    print "Client used unsupported method";
 print LOGFILE "\nMethod found was: Client used unsupported method\n";
}

我的假设是，在小程序中使用了像这样的一些代码:

import org.apache.commons.httpclient.HttpClient;
import org.apache.commons.httpclient.HttpStatus;
import org.apache.commons.httpclient.methods.PostMethod;

import java.io.BufferedReader;
import java.io.InputStreamReader;

public class PostMethodExample {

  public static void main(String args[]) {

    HttpClient client = new HttpClient();
    client.getParams().setParameter("http.useragent", "Test Client");

    BufferedReader br = null;

    PostMethod method = new PostMethod("http://search.yahoo.com/search");
    method.addParameter("p", "\"java2s\"");

    try{
      int returnCode = client.executeMethod(method);

      if(returnCode == HttpStatus.SC_NOT_IMPLEMENTED) {
        System.err.println("The Post method is not implemented by this URI");
        // still consume the response body
        method.getResponseBodyAsString();
      } else {
        br = new BufferedReader(new InputStreamReader(method.getResponseBodyAsStream()));
        String readLine;
        while(((readLine = br.readLine()) != null)) {
          System.err.println(readLine);
      }
      }
    } catch (Exception e) {
      System.err.println(e);
    } finally {
      method.releaseConnection();
      if(br != null) try { br.close(); } catch (Exception fe) {}
    }

  }
}

Answer 1

你不能只使用Request.Form吗？而不是 Request.QueryString ？

例如:

string p = Request.Form["p"];    // should contain "java2s" in your example