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1 | initial version |
Normally I open the shell of Ubuntu and write
ssh ubuntu@ubiquityrobot.local
. Then I have to write the password and finally I can access my raspberry PI. I would like to make the same procedure with my webpage but without using the shell. Therefore I think that I need this functionmainSocket.onopen = ...
I understand. Based on the above information, I need to make the following points:
ssh
access to the Raspberry PI using a WebSocket
? Please think about it!rosbridge_server
and ssh
runs on 9090 and 22 port, respectively. Please check and correct those port numbers.ssh
connection using a WebSocket
. Moreover, I do not understand why someone would pick a WebSocket
to connect to an ssh
server. Please read a similar question asked in StackOverflow for further details.2 | No.2 Revision |
First, you are connecting to rosbridge_server
at ws://192.168.100.34:8080
, and then you are creating a WebSocket
at the same URI. Later you seem to send data to that socket. This is probably the cause of the error. You should not touch the URI for rosbridge_server
again. Furthermore, the default port for rosbridge_server
is 9090. If I remember correctly, port 8080 was used by tomcat in the past. So, I would not pick this port number for rosbridge_server
.
Normally I open the shell of Ubuntu and write
ssh ubuntu@ubiquityrobot.local
. Then I have to write the password and finally I can access my raspberry PI. I would like to make the same procedure with my webpage but without using the shell. Therefore I think that I need this functionmainSocket.onopen = ...
I understand. Based on the above information, I need to make the following points:
ssh
access to the Raspberry PI using a WebSocket
? Please think about it!rosbridge_server
and ssh
runs on 9090 and 22 port, respectively. Please check and correct those port numbers.ssh
connection using a WebSocket
. Moreover, I do not understand why someone would pick a WebSocket
to connect to an ssh
server. Please read a similar question asked in StackOverflow for further details.