Loading GZIP InputStream Into XML Doc RSS

• • Reply
  kevorkian

  Member

  301 Points

  393 Posts

  Loading GZIP InputStream Into XML Doc

  Aug 31, 2015 06:19 PM|kevorkian|LINK

  I have a page that accepts XML, and I've always loaded it for parsing by simply doing: 

  Dim xDoc As XmlDocument = New XmlDocument()
  xDoc.Load(Request.InputStream)

  However the input stream is coming over as a string of digits, like so:
  
  60631201091083211810111411510511111061344....(many, many more)
  
  At first I thought it was base64, but that isn't the case.  How do I decode this?

• • Reply
  cnranasinghe

  Contributor

  6711 Points

  2334 Posts

  Re: Loading GZIP InputStream Into XML Doc

  Aug 31, 2015 11:08 PM|cnranasinghe|LINK

  Hope you can adapt the below code for  your requirement, code below turns an object into xml document and put it on memory stream.

  on the click event

    Dim currentStream As New MemoryStream()
    Dim streamWriter As New StreamWriter(currentStream)
    streamWriter.Write(CreateXMLDoc(reportDetail))
  
    streamWriter.Flush()
    currentStream.Position = 0
  

  Now the "currentStream" contains xml document content as string, Here "reportDetail" is the object which needed to converted to xml. alter this function as to your requirement.

      Public Shared Function CreateXMLDoc(ByVal value As Object) As String
          Dim rv As String = ""
          Dim xml = New XmlSerializer(value.[GetType]())
          Using memoryStream As MemoryStream = New MemoryStream()
  
              Using xmlWriter As XmlTextWriter = New XmlTextWriter(memoryStream, Encoding.UTF8)
                  xmlWriter.Formatting = Formatting.Indented
                  xml.Serialize(xmlWriter, value)
                  rv = Encoding.UTF8.GetString(memoryStream.ToArray())
              End Using
  
          End Using
          Return rv
      End Function

  you can use the currentStream variable to manipulate the xml,  like sending an email with xml attachment.

• • Reply
  Weibo Zhang

  Star

  7980 Points

  1586 Posts

  Re: Loading GZIP InputStream Into XML Doc

  Sep 01, 2015 03:29 AM|Weibo Zhang|LINK

  Hi kevorkian,

  You could first save the “Request.InputStream” to check whether the content is right.  Besides, you could try to use “Request.Files[0].InputStream” to get the HttpPostedFile. If the issue still exists, you could provide the whole code about how to save the xml files to Steam.

  Dim stream As New StreamReader(Request.Files(0).InputStream)
  Dim x As String = stream.ReadToEnd()
  Dim xml As String = HttpUtility.UrlDecode(x)
  Dim document = New XmlDocument()
  document.LoadXml(xml) 
  

  https://msdn.microsoft.com/en-us/library/system.web.httppostedfile.inputstream(v=vs.110).aspx

  I hope it’s useful to you.

  Best Regard,

  Weibo Zhang

• • Reply
  kevorkian

  Member

  301 Points

  393 Posts

  Re: Loading GZIP InputStream Into XML Doc

  Sep 02, 2015 10:04 PM|kevorkian|LINK

  Here is the post header:
  
  

  Header Info: Cache-Control=no-cache

  Connection=close

  Content-Length=413

  Content-Type=application/xml

  Accept=text/xml, application/xml, text/html, text/plain, */* Accept-Encoding=gzip, deflate, * Accept-Language=en-us, en, * Authorization=Basic Og== Host=myhost.com

  User-Agent=Mozilla/4.0 (compatible; MSIE 6.0; Windows 98; Win 9x 4.90; Q312461)
  
  The only method I have found effective to obtain anything from the inputstream, is this:
  
  

              Str = Request.InputStream
              strLen = CInt(Str.Length)
              Dim strArr(strLen) As Byte
              strRead = Str.Read(strArr, 0, strLen)
  
              For counter = 0 To strLen - 1
                  strmContents = strmContents & strArr(counter).ToString()
              Next counter

  
  The contents of strmContents is:
  
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

• • Reply
  Weibo Zhang

  Star

  7980 Points

  1586 Posts

  Re: Loading GZIP InputStream Into XML Doc

  Sep 13, 2015 07:35 AM|Weibo Zhang|LINK

  Hi kevorkian,

  kevorkian

  The only method I have found effective to obtain anything from the inputstream, is this:

  Did the method I provide you above have some issue on your client? Could you show the result of the above method? If the method also has the same issue, could you show how you accept the XML? Besides, you could also take a look at the following similar thread which shows how to get the xml file string.

  http://stackoverflow.com/questions/10049003/retrieve-http-post-request-containing-xml

  I hope it’s useful to you.

  Best Regards,

  Weibo Zhang

• • Reply
  kevorkian

  Member

  301 Points

  393 Posts

  Re: Loading GZIP InputStream Into XML Doc

  Sep 16, 2015 04:54 PM|kevorkian|LINK

  Ok, so still can't figure this out.  Here is the post header:
  
  Cache-Control=no-cache Connection=close Content-Length=413 Content-Type=application/xml Accept=text/xml, application/xml, text/html, text/plain, */* Accept-Encoding=gzip, deflate, * Accept-Language=en-us, en, * Authorization=Basic Og== Host=myhost.com User-Agent=Mozilla/4.0 (compatible; MSIE 6.0; Windows 98; Win 9x 4.90; Q312461) 

  Here is how I am attempting to decode the stream:
  
  

   ' Create a Stream object.
              Str = Request.InputStream
              ' Find number of bytes in stream.
              strLen = CInt(Str.Length)
              ' Create a byte array. 
              Dim strArr(strLen) As Byte
              ' Read stream into byte array.
              strRead = Str.Read(strArr, 0, strLen)
  
              ' Convert byte array to a text string. 
              For counter = 0 To strLen - 1
                  strmContents = strmContents & strArr(counter).ToString()
              Next counter
  
  
              Dim textResponse = Decompress(strArr)

  Private Shared Function Decompress(data As Byte()) As Byte()
          Using compressedStream = New MemoryStream(data)
              compressedStream.Seek(0, SeekOrigin.Begin)
              Using zipStream = New GZipStream(compressedStream, CompressionMode.Decompress)
                  Using resultStream = New MemoryStream()
                      zipStream.CopyTo(resultStream)
                      Return resultStream.ToArray()
                  End Using
              End Using
          End Using
      End Function

  That function results in the exception:
  
  The magic number in GZip header is not correct. Make sure you are passing in a GZip stream.