Hi all I’m trying to send a file using the windows sockets API’s and this is how I’m doing it: 'code for VB6 'first loading the file to a string buffer. Open "File Name " For Binary As #1 Dim FileData As String FileData = Space$(LOF(1)) Get #1, , FileData Close #1 'now sending the file Dim SendBuffer() As Byte FileData = FileData & "FileEnd" SendBuffer() = StrConv(FileData, vbFromUnicode) Call send(Socket, Buffer(0), Len(FileData), 0&) DoEvents 'I check if the file has arrived using the following routine Public FileBuffer As String Dim TempBuffer As String Dim BytesReceived As Long Dim ByteBuffer(1 To 8192) As Byte BytesReceived = recv("The Socket Handel", _ ByteBuffer(1), _ 8192, 0&) If BytesReceived > 0 Then TempBuffer = StrConv(ByteBuffer, vbUnicode) If Right(TempBuffer, 7) = "FileEnd" Then FileBuffer = FileBuffer & Left(TempBuffer, Len(TempBuffer) - 7) Open "FileName" For Binary As #1 Put #1, , FileBuffer Close #1 Else FileBuffer = FileBuffer & TempBuffer End If End If The problem starts if I try to send anything through the same socket right after sending the file "For Example the file size" because the condition : if right(TempBuffer,7)="FileEnd" will always be false because the (TempBuffer =”The File data” and what ever information I send right after sending the file). I tried to solve this problem using the “sleep( )” function after sending the file data, but it is not reliable. Is there a better way to do this without using another socket or using a long period for the sleep( ) function? Any help is greatly appreciated. Thanks all. [edited by - Fma on July 2, 2002 6:44:38 AM]