Timeline for List of connected clients to server
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jan 3, 2017 at 21:57 | comment | added | Nick Gammon♦ | @MarkSmith - the question was How can I see if this client connected already - well, if the client is connected they are in the list of connected clients. To answer the other part of the question a simple array would probably do, as you can only have about 4 connections on Ethernet shields anyway. | |
Jan 3, 2017 at 21:40 | comment | added | Mark Smith | I still don't understand what that's got to do with anything here, but perhaps it doesn't matter :-) | |
Jan 3, 2017 at 21:34 | history | edited | Nick Gammon ♦ | CC BY-SA 3.0 |
Added more explanations.
|
Jan 3, 2017 at 21:32 | comment | added | Nick Gammon♦ | @MarkSmith - well, if you are on the phone to Aunt Ruth, you know you are talking to her, right? You don't need to write it down on a list. So if you have a number of connections open, they are in a list somewhere already, otherwise you couldn't be using them. | |
Jan 3, 2017 at 21:31 | comment | added | Nick Gammon♦ | Ah, OK. But you can still choose whether or not they are persistent. And judging by what I read some servers time out fairly quickly (15 seconds). | |
Jan 3, 2017 at 17:44 | comment | added | SoreDakeNoKoto |
+1 I think you meant to say Under HTTP 1.0... since HTTP 1.1 connections are persistent by default.
|
|
Jan 3, 2017 at 9:47 | comment | added | Mark Smith | I'm not sure I understand what you mean. | |
Jan 3, 2017 at 8:27 | comment | added | Nick Gammon♦ | Good point. Still, if you have a number of connections from TCP sockets, surely you know what they are already? | |
Jan 3, 2017 at 8:15 | comment | added | Mark Smith | Good explanation of HTTP. Note however that the OP doesn't mention using HTTP ☺️. It's perfectly valid to have a normal TCP socket connection open and for data to trickle down it as and when the server feels like it. | |
Jan 3, 2017 at 2:00 | history | answered | Nick Gammon ♦ | CC BY-SA 3.0 |