B4J Question How do I get the content of multipart/form-data?

[hzq200409]

For the following requests in jServer, is there a quick way to get them? For example, user_name.

--------------------------00077618139f9eb3
Content-Disposition: form-data; name="channel_num"

1
--------------------------00077618139f9eb3
Content-Disposition: form-data; name="ipaddr"

10.2.50.5
--------------------------00077618139f9eb3
Content-Disposition: form-data; name="port"

8000
--------------------------00077618139f9eb3
Content-Disposition: form-data; name="user_name"

temp
--------------------------00077618139f9eb3--

 

[peacemaker]

Maybe back-coding of this ?

[B4X] Post multipart requests / file uploads with progress

This code will work in B4A and B4J. It will not work in B4i. Public Sub CreateMultipartJob(Link As String, NameValues As Map, Files As List) As HttpJob Dim boundary As String = "---------------------------1461124740692" TempCounter = TempCounter + 1 Dim TempFileName As String =...

www.b4x.com

 

[hzq200409]

Maybe back-coding of this ?

[B4X] Post multipart requests / file uploads with progress

This code will work in B4A and B4J. It will not work in B4i. Public Sub CreateMultipartJob(Link As String, NameValues As Map, Files As List) As HttpJob Dim boundary As String = "---------------------------1461124740692" TempCounter = TempCounter + 1 Dim TempFileName As String =...

www.b4x.com

Thank you for your attention. This post is about uploading. As an aside: and this upload example is not suitable for all situations, I have modified this code in a similar use environment.

 

[hzq200409]

I'm writing the code to get them right now, which is a bit of a hassle.

 

[peacemaker]

How can I See the content of a MultipartForm request?

I am using Apache HTTPClient 4. I am doing very normal multipart stuff like this: val entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE); entity.addPart("filename", new FileBody(ne...

stackoverflow.com

 

[hzq200409]

How can I See the content of a MultipartForm request?

I am using Apache HTTPClient 4. I am doing very normal multipart stuff like this: val entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE); entity.addPart("filename", new FileBody(ne...

stackoverflow.com

Sorry, my English is not good enough. You may have misunderstood my meaning. I am writing the jServer program and I want to extract them in the service program.

 

[hzq200409]

    If req.Method = "POST" Then
        If req.ContentType.Contains("multipart/form-data") Then
            Dim partData As Map = req.GetMultipartData(File.DirApp & "/tempUpload",2048000)
            If partData.IsInitialized Then
                For Each k As String In partData.Keys
                    Dim p As Part = partData.Get(k)
                    Dim Value As String = p.GetValue("utf8")
                    Log(k & ":" & Value)
                Next
            End If
        End If
    End If

Dim content As TextReader
content.Initialize(req.InputStream) <----Wrong usage. Because the existence of this code makes the above code invalid. I fell into this trap.Because this code comes before the code above.