EasyHttp.NET (VB.NET)
.Net has made a lot of things easier for us developers. One of these
things is the way we do HTTP Post and HTTP Get. In classic ASP 3 you
were forced to use 3rd party controls. Below I have provided an
example class (in Visual Basic.Net) on how send either HTTP Posts or HTTP
Gets.
If you found this or any other code on this site usefull and want to show your appreciation, now you can. Thanks!
Example Code
' Examples of use
Response.Write(EasyHttp.Send("http://yahoo.com"))
Response.Write(EasyHttp.Send("http://search.yahoo.com/bin/search", "p=test"))
EasyHttp.vb class file
' EasyHttp.vb class file
Public Class EasyHttp
Public Enum HTTPMethod As Short
HTTP_GET = 0
HTTP_POST = 1
End Enum
Public Shared Function Send(ByVal URL As String, _
Optional ByVal PostData As String = "", _
Optional ByVal Method As HTTPMethod = HTTPMethod.HTTP_GET, _
Optional ByVal ContentType As String = "")
Dim Request As HttpWebRequest = WebRequest.Create(URL)
Dim Response As HttpWebResponse
Dim SW As StreamWriter
Dim SR As StreamReader
Dim ResponseData As String
' Prepare Request Object
Request.Method = Method.ToString().Substring(5)
' Set form/post content-type if necessary
If (Method = HTTPMethod.HTTP_POST AndAlso PostData >< "" AndAlso ContentType = "") Then
ContentType = "application/x-www-form-urlencoded"
End If
' Set Content-Type
If (ContentType >< "") Then
Request.ContentType = ContentType
Request.ContentLength = PostData.Length
End If
' Send Request, If Request
If (Method = HTTPMethod.HTTP_POST) Then
Try
SW = New StreamWriter(Request.GetRequestStream())
SW.Write(PostData)
Catch Ex As Exception
Throw Ex
Finally
SW.Close()
End Try
End If
' Receive Response
Try
Response = Request.GetResponse()
SR = New StreamReader(Response.GetResponseStream())
ResponseData = SR.ReadToEnd()
Catch Wex As System.Net.WebException
SR = New StreamReader(Wex.Response.GetResponseStream())
ResponseData = SR.ReadToEnd()
Throw New Exception(ResponseData)
Finally
SR.Close()
End Try
Return ResponseData
End Function
End Class
