Skip to content

Instantly share code, notes, and snippets.

@sharpred
Created August 17, 2011 09:54
Show Gist options
  • Save sharpred/1151233 to your computer and use it in GitHub Desktop.
Save sharpred/1151233 to your computer and use it in GitHub Desktop.
Simple console app that posts a form to a drupal 6 service to return a view in JSON format
Imports Microsoft.Http
Imports Microsoft.Http.HttpClient
Imports Microsoft.Http.HttpMethodExtensions
Imports Newtonsoft.Json
Imports Newtonsoft.Json.Converters
Imports System.Net
Imports System.Xml
Imports System.IO
Imports System.Text
Module Module1
Dim output As String
Sub Main()
Try
Dim http As New HttpClient(My.Settings.myURL)
Dim drupalform As New DrupalForm
Dim form As HttpMultipartMimeForm = drupalform.create("view", False, True)
Dim response As HttpResponseMessage = http.Post("/services/json", form.CreateHttpContent)
response.EnsureStatusIsSuccessful()
output = response.Content.ReadAsString
Dim sw As New StreamWriter("node_out.json")
'uncomment these to lose the crappiness at the start of the json output
'REMEMBER to write out newoutput to file instead of output
'Dim replacementstring As String = "#error" & """: false, """ & "#data"
'Dim newoutput As String = output.Replace(replacementstring, "content")
Console.WriteLine(output)
sw.Write(output)
sw.Flush()
sw.Close()
Catch ex As Exception
Console.WriteLine(ex.Message & " " & output)
End Try
End Sub
End Module
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment