1

哇，我已经尝试了从编写自己的标头到转换 C# 示例的所有方法，但在上传照片时我仍然无法通过 400 Bad Request 错误。

我添加了所有可能的权限，并且我的令牌是正确的。

我可以将状态更新发布到我的提要中，但我无法上传图片。这是我尝试过的两种不同方法，都给了我 400 Bad Request ...

1

    Dim myReq As HttpWebRequest
    Dim myRes As HttpWebResponse
    Dim encoding As New System.Text.ASCIIEncoding()
    Dim postData As String
    Dim data() As Byte
    Dim sr As StreamReader
    Dim imagedata As String

    imagedata = File.OpenText("C:\ebay00042-1.jpg").ReadToEnd()

    postData += "access_token=MY_TOKEN_HERE_29ZB51pPizthxX5lhmst3MZC7hYXQhW8ZB8e7sVVLzEaN8ZCZAzAgrzk1pisw3ZCtK5lwMMTZBUhe07xTsQvfeHosA1GFUAZDZD&message=this is a test123&source=" & imagedata 'File.ReadAllBytes(photoPath)
    data = encoding.GetBytes(postData)

    myReq = WebRequest.Create("https://graph.facebook.com/380406275386560/photos")
    DirectCast(myReq, System.Net.HttpWebRequest).UserAgent = "Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1)"
    myReq.Method = "POST"
    myReq.ContentType = "application/x-www-form-urlencoded"
    myReq.ContentLength = data.Length
    Dim myStream As Stream = myReq.GetRequestStream
    myStream.Write(data, 0, data.Length)
    myStream.Close()
    myRes = myReq.GetResponse
    sr = New StreamReader(myRes.GetResponseStream)
    Dim strHTML As String = sr.ReadToEnd

2，尝试创建自己的标题..

    Dim myReq As HttpWebRequest
    Dim myRes As HttpWebResponse
    Dim encoding As New System.Text.ASCIIEncoding()
    Dim data() As Byte
    Dim sr As StreamReader
    Dim boundary As String = "----------" + DateTime.Now.Ticks.ToString("x")


    Dim sb As StringBuilder = New StringBuilder("")
    sb.Append("----------").Append(boundary).Append("\r\n")
    sb.Append("Content-Disposition: form-data; name=""access_token""").Append("\r\n")
    sb.Append("\r\n")
    sb.Append("MY_TOKEN_HERE_MZC7hYXQhW8ZB8e7sVVLzEaN8ZCZAzAgrzk1pisw3ZCtK5lwMMTZBUhe07xTsQvfeHosA1GFUAZDZD").Append("\r\n")

    sb.Append("----------").Append(boundary).Append("\r\n")
    sb.Append("Content-Disposition: form-data; name=""message""").Append("\r\n")
    sb.Append("\r\n")
    sb.Append("Testttt").Append("\r\n")

    sb.Append("----------").Append(boundary)
    sb.Append("Content-Disposition: file; name=""source"" filename=""ebay00042-1.jpg""").Append("\r\n")
    sb.Append("Content-Type: image/jpeg).Append(\r\n")
    'sb.Append("Content-Transfer-Encoding: binary").Append("\r\n")
    sb.Append("\r\n")
    sb.Append(File.OpenText("C:\ebay00042-1.jpg").ReadToEnd()).Append("\r\n")

    sb.Append("----------").Append(boundary).Append("----------").Append("\r\n")


    'txtCaption.Text = sb.ToString
    data = encoding.GetBytes(sb.ToString)


    myReq = WebRequest.Create("https://graph.facebook.com/380406275386560/photos")
    DirectCast(myReq, System.Net.HttpWebRequest).UserAgent = "Mozilla/4.0 (compatible; MSIE 7.0; Windows NT 5.1)"
    myReq.Method = "POST"
    myReq.ContentType = "multipart/form-data; boundary=" + boundary
    'myReq.ContentType = "application/x-www-form-urlencoded"
    myReq.ContentLength = data.Length
    Dim myStream As Stream = myReq.GetRequestStream
    myStream.Write(data, 0, data.Length)
    myStream.Close()
    myRes = myReq.GetResponse
    sr = New StreamReader(myRes.GetResponseStream)
    Dim strHTML As String = sr.ReadToEnd

任何帮助将不胜感激！

4

1 回答 1

0

我对 Facebook 的 API 一无所知，但如果我是你，我会好好看看http://csharpsdk.org/

如果我必须根据您提供的信息发送图像，它看起来像这样：

' Request URL, image file to send, token and result HTML buffer
Dim reqUrl As String = "https://graph.facebook.com/380406275386560/photos"
Dim imageData As Byte() = File.ReadAllBytes("C:\ebay00042-1.jpg")
Dim token As String = "MY_TOKEN_HERE"
Dim strHtml As String = ""
' Request
Dim request As WebRequest = WebRequest.Create(reqUrl)
request.Headers.Add("access_token", token)
request.Method = "POST"
' set *correct* content type
request.ContentType = "image/jpeg"
' write image data to request stream
Using str = request.GetRequestStream()
    str.Write(imageData, 0, imageData.Length)
End Using
' response
Dim response As WebResponse = request.GetResponse()
' HTTP Status
Dim status As Integer = CType(response, HttpWebResponse).StatusCode
If status = 200 Then
    ' success
    Using reader As New StreamReader(response.GetResponseStream())
        strHtml = reader.ReadToEnd()
    End Using
Else
    ' oops
End If

希望有帮助。

于 2012-12-16T00:49:48.223 回答