Я ищу решение проблемы. Это сообщение в виде скручиваемости в качестве ссылки для выяснения данных json на платформе Favoriot.com.
curl -X POST --header 'Content-Type: application/json'
--header 'Accept: application/json'
--header 'apiKey: YOUR API KEY HERE'
-d '{
"device_developer_id": "deviceDefault@FAVORIOT",
"data": { "temperature": "31","humidity": "70"}
}'
Как я уже упоминал выше, при попытке публикации с использованием vb net .. Я добавил исходный код в качестве ссылки.
Imports Newtonsoft.Json
Imports System.Net
Imports System.Text
Public Class Form1
Private urlToPost As String = ""
Private apikey_favoriot As String = "YOUR API KEY HERE"
Public Sub New(ByVal urlToPost As String)
Me.urlToPost = urlToPost
End Sub
Public Function postData(ByVal dictData As Dictionary(Of String, Object)) As Boolean
Dim webClient As New WebClient()
Dim resByte As Byte()
Dim resString As String
Dim reqString() As Byte
Try
webClient.Headers("content-type") = "application/json"
webClient.Headers.Add("apikey", apikey_favoriot)
reqString = Encoding.Default.GetBytes(JsonConvert.SerializeObject(dictData, Formatting.Indented))
resByte = webClient.UploadData(Me.urlToPost, "post", reqString)
resString = Encoding.Default.GetString(resByte)
Console.WriteLine(resString)
webClient.Dispose()
Return True
Catch ex As Exception
Console.WriteLine(ex.Message)
End Try
Return False
End Function
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
Dim Form1 As New Form1("http://192.168.254.104:8000")
Dim dictData As New Dictionary(Of String, Object)
dictData.Add("device_developer_id", "deviceDefault@FAVORIOT")
dictData.Add("data", "temperature : 31")
Form1.postData(dictData)
End Sub
End Class