Bash Script, lsusb

[conradin]

Hi all I have a usb device that I want to poll periodically.
the device is listed as:

lsusb | grep "Tiger Jet Network"
Bus 002 Device 004: ID 06e6:c200 Tiger Jet Network, Inc. Internet Phone

all I want to know is when its plugged in or not plugged in.
how might I do that?

Im just starting to read about polling, lsusb tells me what I want to know, but I dont know how to put it into a true false statement in a bash script.

#!/bin/bash

result=0
result=`lsusb | grep -e "Tiger Jet Network, Inc. Internet Phone"`
if $result !=0
then
echo "Great execute some code"
else
echo "Device was unpluged"
exit
fi

That is what I had tried thus far, but it doesnt work.
can someone point me in the direction of being able to execute code if
lsusb says my device is plugged in?

[drmrgd]

Not sure if this is exactly correct, but I think you can just use the status of the shell exit code in that conditional statement (also have a look at your conditional statement construct...not sure that's quite correct):


#!/bin/bash

result=$(lsusb | grep -e "Tiger Jet Network, Inc. Internet Phone")

if [ $? -ne 0 ]; then
printf "Device is unplugged\n"
else
printf "Great execute some code\n"
fi

Give that a shot to see if it works as you expect

[conradin]

That does work exactly as I want, Thank you very much!
I think I have a better understanding of $ now too.
But there are persistently things I dont understand about $. Like, what is the meaning of "$?" ..?




Not sure if this is exactly correct, but I think you can just use the status of the shell exit code in that conditional statement (also have a look at your conditional statement construct...not sure that's quite correct):


#!/bin/bash

result=$(lsusb | grep -e "Tiger Jet Network, Inc. Internet Phone")

if [ $? -ne 0 ]; then
printf "Device is unplugged\n"
else
printf "Great execute some code\n"
fi

Give that a shot to see if it works as you expect

[Lars Noodén]

There are special variables that contain various pieces of useful information. $? provides the exit status of the last program run. $$ is the process ID of the current script.

Scroll down this page to see more:
http://tldp.org/LDP/abs/html/internalvariables.html

[sisco311]

There are special variables that contain various pieces of useful information. $? provides the exit status of the last program run. $$ is the process ID of the current script.

Scroll down this page to see more:
http://tldp.org/LDP/abs/html/internalvariables.html

That link contains inaccurate informations. $BASH or $whatever is not an internal variable or builtin variable. It's a parameter expansion. BASH is a shell variable and whatever is a parameter :)

See:

man bash | less +/" + Parameter Expansion"

$ is the expansion character. It's a special character (http://mywiki.wooledge.org/BashGuide/SpecialCharacters) or metacharacter used in substitutions: parameter and variable expansion, command substitution and arithmetic expansion.


? and $ (yep, whatever includes the $ sign too :) ) are special parameters (http://mywiki.wooledge.org/BashGuide/Parameters).

$? expands to the exit code of the most recently completed foreground command.

$$ expands to the PID (process ID number) of the current shell. The first $ sign is a the expansion character and the second is the special parameter.

[Lars Noodén]

That link contains inaccurate informations.

Thanks. The e-mail address of the author can be found on this link: http://tldp.org/LDP/abs/html/
Could you report the mistake to him?

In perl, these are actual special variables Possibly it is the same in other languages and part of the mix up.