i want to change the open dialog box that is create by MFC,
how can i make the changes to the open dialog box that users ONLY can view the folders and files from the server, so that user just can open the files from that server.
thanks for reply, what i want is when you press the open button from file menu, it will prompt a open dialog box, right. Basically that box can open the files from c or d drives, but now i want that open dialog box display the ftp server files, not the files from the drives.
I wrote Socket programming which is running good, i want to know whether all the data is received properly by the server with out any loss from client side. Meaning i need acknowledge from server side. How can i do it?.
All IP-packets are sent with a CRC32 checksum that the receiver recalculates to validate the data. If the checksum is incorrect, the packet is discarded.
This happens regardless of what you do, it's how the IP-protocol works.
If you're using an UDP ports, the sender will never be informed whether the receiver got the packet or not, at any OSI level, since UDP traffic is connectionless.
If you're using TCP ports, the sender will be sent an ACK for each packet. Whether the packet is to be resent or not is depending on a configurable threshold in the registry found at HKEY_LOCAL_MACHINE/SYSTEM/currentcontrolset/services/tcpip/parameters.
This means that the sender cannot be informed if the receiver got a corrupt packet since the receiving application will never get the packet since it's been discarded.
I suggest you implement a small protocol of your own at the OSI application level that send ACKs to the sender's application. You will then be informed if the receiver never got the packet by the absence of an ACK.
Hope this helps
It's supposed to be hard, otherwise anybody could do it!
Regarding CodeProject: "resistance is pointless; you will be assimilated"