Question

我有一个恼人的ASP.NET问题：

我有一个Perl脚本（见下文），它获取了form_info变量。现在不幸的是，它是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/ app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 ... 我无法将http POST更改为HTTP get，因为它是由第三方java applet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，这样的代码在applet中使用：

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代替Request.QueryString吗？

例如：

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

Answer 2

不，request.form不起作用。首先，我没有名字，索引一和二存在，但没有进一步......

但我刚刚发现了整个周末没有发现的东西：它的工作原理是读取stdin，这相当于在asp.net中读取http头文件

我已经看到了一个asp.net http post请求的示例，并通过使用请求者对象的GetRequest方法读取流来找到答案。我没有任何请求者对象，但我意识到它可能在页面对象中的某个位置。由于搜索到的对象是GetRequest。

，因此逻辑位置被接合为请求

    Function readme() As String
    Dim sr As System.IO.StreamReader = New System.IO.StreamReader(Page.Request.InputStream())
    Return sr.ReadToEnd().Trim()
End Function


Sub WriteToFile(Optional ByRef strStringToWrite As String = "Hello World")
    Dim fp As System.IO.StreamWriter

    Try
        fp = System.IO.File.CreateText(Server.MapPath("./") & "test.txt")
        fp.WriteLine(strStringToWrite)
        Response.Write("File Succesfully created!")
        fp.Close()
    Catch err As Exception
        Response.Write("File Creation failed. Reason is as follows " + err.ToString())
    Finally

    End Try
End Sub

Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load
    'Response.Write(readme)


    WriteToFile(readme())

    'Dim p As String = Request.Form(0)
    'WriteToFile(p)

    'p = Request.Form(2)
    'WriteToFile(p)
End Sub

使用asp.net进行http POST

2 个答案: