Sending data from LoPy to a webserver

[this.wiederkehr]

I'd do it the other way round: Connect the lopy to your local wlan.

you might use the urequests lib from here https://github.com/micropython/micropython-lib/tree/master/urequests

Then use e.g. http post requests to send the data:

import urequests
sensorreading = 'whatever'
res = urequests.post('http://yourserver/endpoint', json={'my_sensor': sensorreading})
res.close()
if (res.status_code - 200) < 100: # check that response code is in the 200 range
    print('done')
else:
    print('there was some error')

This will send a post http request to the given url and the sensor reading as json.

If your are doing this over wan or on an untrusted wlan you have to consider security!

A completely different approach would be to use mqtt. Example here: https://docs.pycom.io/chapter/tutorials/all/mqtt.html
(needs an mqtt broker)

[JeromeMaquoi]

@this-wiederkehr @livius Ok I understood I could connect my LoPy to the webserver via WiFi. But can I do it without stopping to transmit Wifi with the LoPy?

[this.wiederkehr]

@JeromeMaquoi

I thought it is not possible to use the lopy as both, station(connecting to an existing wlan) and ap (access point providing a wlan) simultaneously... but in fact it "seems" to be possible, tough I have not tested. And some restrictions apply (same channel of ap and sta)

https://www.esp32.com/viewtopic.php?t=1192#p5285
https://docs.pycom.io/chapter/firmwareapi/pycom/network/wlan.html

setting the mode as to WLAN.STA_AP seems to make this possible ...

[JeromeMaquoi]

@this-wiederkehr Hello again,

I used the urequests lib that you gave me a month ago. Unfortunately I have this error. Can you help me?

Here is the 24 line:

Also, I have another question. How can I catch those data from my web server?

Best regards,

Jérôme Maquoi

[this.wiederkehr]

127.0.0.1 refers to localhost. you have to use the real ip of the server.

See the example on how to get the response. https://github.com/micropython/micropython-lib/blob/master/urequests/example_xively.py

[JeromeMaquoi]

@this-wiederkehr Hello,
I tested the code with the real ip address and it works! But I don't understand how it works. When I make "urequests.post", where is the data sent? And how can I catch it from the server side? I did'nt understand your response about that.

Jérôme Maquoi

[JeromeMaquoi]

Someone can help me?

[livius]

@jeromemaquoi
real sample tell more
this is the same as when you open your browser and put address in the bar
when you write http://www.pycom.io (if you write only www.pycom.io browser assume http:// at start and also try https)
when you press enter browser send request to the server on default port which is 80 for http and 443 for https
name www.pycom.io is translated to ip 18.220.239.183

---

now "same" is for urequests.post
if you write

urequests.post("http://123.123.123.123/", data = data)

request is send to destination server 123.123.123.123 to port 80
and waiting for response from that server.
If on that server there is nothing listening on port 80 then you got timeout

then to recive something you must have on destination server socket which listen on port 80

[this.wiederkehr]

Your question is very broad ....

The above code sends a post request with the sensor data as a json payload. Each server framework provides a way to access that data however the way to do it depends on the framework.

I see that you have used php, as I have no idea about php I'm quoting stackoverflow for a way to access json in the payload with php:

$data = json_decode(file_get_contents('php://input'), true);

https://stackoverflow.com/a/18867369/4273834

[JeromeMaquoi]

@livius Hi, thanks for the answer. It's clearer for me now.
So if I understood, when I run this code and I have the word "done" printed in the console, does that mean that the server is listening on port 80? Or does that mean that the data is just correctly sent?

[livius]

@jeromemaquoi
yes, you got 'done' if on the destination side port is open and something is listen