Goal: Being able to target an host with a slave and get back all its exported ressources, whatever the protocol. I join Alex in the opinion that in a LAN environment, you might want to check what's on a specific host, independantly of the protocol (nfs, smb, ftp, http,...) More, you might want to get a list of all the machines present on your LAN so as to be able to browse them later on. Problem: For a small LAN, you might get a list of all hosts that offer a specific service, but how would you know which machines offer, for example, ftp or nfs, without scanning them all? It's not acceptable if your LAN gather several hundred machines. Possible solution: Create a virtual protocol, called 'any', that would merge results from different protocols, and can only work when you target a specific host. Of course, one might argue that smb:// is relevant, as it can return a list of workgroups, but what for ftp://, or nfs://? (I definitely don't like this concept of neighborhood. What is considered to be in a machine neighborhood anyway?) So: - 'any://this_host/' would return the list of ressources independantly of the protocol (still problem here on how to merge html pages with nfs shares, but let's assume it returns a list of files & dirs for now). Ressources are then acceses with paths as in any other URL. (any://host/path.../a_file) - 'any://' not valid, unless it is restricted to 'smb://' (except if someone comes up with a solution for other protocols). In this restricted case, you can still get a .desktop link to any:// on your desktop if you really wish so. - You can call 'any' for a non-LAN host as well. - We keep all the existing well-separated kio_slaves in case the user knows what she/he wants. Note about the merging process: Can scan the ports of target host to found supported protocols, and then call the corresponding ioslaves. Conflicts can be handled by prefix depending on the protocol (see knetmon/netknife) and/or setting up a user prefered order. Comments? -- Life is a sexually transmitted fatal disease. (W. Allen?)