WIZnet Developer Forum

End header characters

Please tell me is it possible to somehow avoid terminating 0D and 0A after message:
Reported length is 6, and real length is 8.
I tried with AT+SFORM=111111100,… , but I got even worse situation:
Can message be finished with nothing more, beyond the reported length in header ???


Hi, djdjdjole

As you know,the last two words(0d0a) means crlf(new line).
Therefore, when you tried ‘AT+SFORM=111111100’, there was no crlf.

Would you check what i understand is correct, do you want to do crlf without 0d0a?


Maybe I was not so clear. Default way of coding data (when module is in TSN mode) is to send header info like this: {1,,51310,6}, before sending the message “ABCDEF”. But in reality there is 0D0A attached afther the message - which means: {1,,51310,6}ABCDEF{0D}{0A}. I don’t know what is the meaning of adding 0D0A at the end and also, why it is not reported as overall message length.
What I tried was to find the way to move it away from the message tail and considered AT+SFORM as good point to achieve so. However, I didn’t succseed, I achieved to move 0D or 0A from the tail, but not the BOTH of them. When I try moving both of them - AT+SFORM=111111100,… and look at the message coded by WiFi module- it was: {1,,51472,6}ABCDEF{1,,51472,6}.
No matter, I unwillingly re-coded my source to that situation (0D0A at the tail), but don’t know why it is there, when segment length is already given inside paranthesis. I am forced to do communication in command mode (again unwillingly), because of problems regarding transparent mode that I point to you at our previous mails.



As I said before, ODOA means crlf(New line).
So, when you tried ‘AT+SFORM=111111100,…’, the results make sense.
(Because you set ‘00’ at last two bits(0D0A) , there was no crlf.)
From what I know, there is no way to do crlf(new line) without ODOA.


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