Loading GZIP InputStream Into XML Doc

Question

User-186058747 posted

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?

Answer 1

User-37275327 posted

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.

Answer 2

User-219423983 posted

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

Answer 3

User-186058747 posted

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:

6063120109108321181011141151051111106134494648343210111099111100105110103613485847045563463621310601101011191141119710011595100101991051151051111106213106097112112108105999711610511111095105100625655495248604797112112108105999711610511111095105100621310609911211595971121121081059997116105111110951051006257575748564548485050495560479911211595971121121081059997116105111110951051006213106010010199105115105111110951009711610162504849534548574548506047100101991051151051111109510097116101621310601001019910511510511111095116105109101624956585253584956604710010199105115105111110951161051091016213106010010199105115105111110626860471001019910511510511111062131060115116971161171156251496047115116971161171156213106011211110595114101991011051181011006277604711211110595114101991011051181011006213106011211110595118101114105102105101100627360471121111059511810111410510210510110062131060106111989511610510910195118101114105102105101100627360471061119895116105109101951181011141051021051011006213106047110101119114111971001159510010199105115105111110621310

Answer 4

User-219423983 posted

Hi 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

Answer 5

User-186058747 posted

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.