Загрузить файл в file.io, используя метод POST

Question

Я нашел ссылку в SO, которая может иметь значение для этого запроса Загрузить изображение в file.io (HTTP HTTP) в VBA Код по этой ссылке

Sub UploadFilesUsingVBAORIGINAL()
     'this proc will upload below files to https://file.io/
          '  png, jpg, txt

        Dim fileFullPath As String
        fileFullPath = ThisWorkbook.Path & "\Sample.txt"

        POST_multipart_form_dataO fileFullPath
    End Sub

Private Function GetGUID() As String
    ' Generate uuid version 4 using VBA
    GetGUID = WorksheetFunction.Concat(WorksheetFunction.Dec2Hex(WorksheetFunction.RandBetween(0, 4294967295#), 8), "-", WorksheetFunction.Dec2Hex(WorksheetFunction.RandBetween(0, 65535), 4), "-", WorksheetFunction.Dec2Hex(WorksheetFunction.RandBetween(16384, 20479), 4), "-", WorksheetFunction.Dec2Hex(WorksheetFunction.RandBetween(32768, 49151), 4), "-", WorksheetFunction.Dec2Hex(WorksheetFunction.RandBetween(0, 65535), 4), WorksheetFunction.Dec2Hex(WorksheetFunction.RandBetween(0, 4294967295#), 8))

End Function

Private Function GetFileSize(fileFullPath As String) As Long

    Dim lngFSize As Long, lngDSize As Long
    Dim oFO As Object, OFS As Object

    lngFSize = 0
    Set OFS = CreateObject("Scripting.FileSystemObject")

    If OFS.FileExists(fileFullPath) Then
        Set oFO = OFS.GetFile(fileFullPath)
        GetFileSize = oFO.Size
    Else
        GetFileSize = 0
    End If

    Set oFO = Nothing
    Set OFS = Nothing
End Function



Private Function ReadBinary(strFilePath As String)
    Dim ado As Object, bytFile
    Set ado = CreateObject("ADODB.Stream")
    ado.Type = 1
    ado.Open
    ado.LoadFromFile strFilePath
    bytFile = ado.Read
    ado.Close

    ReadBinary = bytFile

    Set ado = Nothing
End Function


Private Function toArray(str)
    Dim ado As Object
     Set ado = CreateObject("ADODB.Stream")
     ado.Type = 2
     ado.Charset = "_autodetect"
     ado.Open
     ado.WriteText (str)
     ado.Position = 0
     ado.Type = 1
     toArray = ado.Read()
     Set ado = Nothing
End Function


Sub POST_multipart_form_dataO(filePath As String)

    Dim oFields As Object, ado As Object
    Dim sBoundary As String, sPayLoad As String, GUID As String
    Dim fileType As String, fileExtn As String, fileName As String
    Dim sName As Variant

    fileName = Right(filePath, Len(filePath) - InStrRev(filePath, "\"))
    fileExtn = Right(filePath, Len(fileName) - InStrRev(fileName, "."))

    Select Case fileExtn
     Case "png"
        fileType = "image/png"
     Case "jpg"
        fileType = "image/jpeg"
     Case "txt"
        fileType = "text/plain"
    End Select

    Set oFields = CreateObject("Scripting.Dictionary")
    With oFields
        .Add "qquuid", LCase(GetGUID)
        .Add "qqtotalfilesize", GetFileSize(filePath)
    End With

    sBoundary = String(27, "-") & "7e234f1f1d0654"
    sPayLoad = ""
    For Each sName In oFields
        sPayLoad = sPayLoad & "--" & sBoundary & vbCrLf
        sPayLoad = sPayLoad & "Content-Disposition: form-data; name=""" & sName & """" & vbCrLf & vbCrLf
        sPayLoad = sPayLoad & oFields(sName) & vbCrLf
    Next

    sPayLoad = sPayLoad & "--" & sBoundary & vbCrLf
    sPayLoad = sPayLoad & "Content-Disposition: form-data; name=""file""; " & "filename=""" & fileName & """" & vbCrLf
    sPayLoad = sPayLoad & "Content-Type: " & fileType & vbCrLf & vbCrLf & vbCrLf & vbCrLf & vbCrLf



     sPayLoad = sPayLoad & "--" & sBoundary & "--"


      Set ado = CreateObject("ADODB.Stream")
      ado.Type = 1
      ado.Open
      ado.Write toArray(sPayLoad)
      ado.Write ReadBinary(filePath)
      ado.Position = 0

    With CreateObject("MSXML2.ServerXMLHTTP")
        .Open "POST", "https://file.io", False
        .setRequestHeader "Content-Type", "multipart/form-data; boundary=" & sBoundary
        .send (ado.Read())
        Debug.Print .responseText
    End With

End Sub

Любой желающий может попробовать этот код, так как сайт бесплатный. Когда я запустил код, я получил «Успех» в ближайшем окне и получил ссылку на загруженный файл. Кажется, с этим нет проблем, но когда я взял ссылку и поместил ее в браузер, я получил страницу 404, которая не найдена

Я попытался загрузить тот же файл вручную, и он работает без проблем, как и для полученной ссылки. из этого руководства шаги

Любая помощь, пожалуйста?

Написал здесь тоже https://chandoo.org/forum/threads/upload-file-to-file-io-using-post-method.43925/

Answer 1

Мне кажется, что окончательная граница находится не в том месте ie перед содержимым файла. Попробуйте

Sub UploadToIO()

    Const PATH = "c:\tmp\"
    Const FILENAME = "testimage.png"
    Const CONTENT = "image/png"
    Const URL = "https://file.io"

    ' generate boundary
    Dim BOUNDARY, s As String, n As Integer
    For n = 1 To 16: s = s & Chr(65 + Int(Rnd * 25)): Next
    BOUNDARY = s & CDbl(Now)

    Dim part As String, ado As Object
    part = "--" & BOUNDARY & vbCrLf
    part = part & "Content-Disposition: form-data; name=""file""; filename=""" & FILENAME & """" & vbCrLf
    part = part & "Content-Type: " & CONTENT & vbCrLf & vbCrLf

    ' read file into image
    Dim image
    Set ado = CreateObject("ADODB.Stream")
    ado.Type = 1 'binary
    ado.Open
    ado.LoadFromFile PATH & FILENAME
    ado.Position = 0
    image = ado.read
    ado.Close

    ' combine part, image , end
    ado.Open
    ado.Position = 0
    ado.Type = 1 ' binary
    ado.Write ToBytes(part)
    ado.Write image
    ado.Write ToBytes(vbCrLf & "--" & BOUNDARY & "---")
    ado.Position = 0
    'ado.savetofile "c:\tmp\debug.bin", 2 ' overwrite

    ' send request
    With CreateObject("MSXML2.ServerXMLHTTP")
        .Open "POST", URL, False
        .setRequestHeader "Content-Type", "multipart/form-data; boundary=" & BOUNDARY
        .send ado.read
        Debug.Print .responseText
    End With

    MsgBox "File: " & PATH & FILENAME & vbCrLf & _
           "Boundary: " & BOUNDARY, vbInformation, "Uploaded to " & URL

End Sub

Function ToBytes(str As String) As Variant

    Dim ado As Object
    Set ado = CreateObject("ADODB.Stream")
    ado.Open
    ado.Type = 2 ' text
    ado.Charset = "_autodetect"
    ado.WriteText str
    ado.Position = 0
    ado.Type = 1
    ToBytes = ado.read
    ado.Close

End Function