Bash Script, lsusb

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

Code:

---

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.

Code:

---

#!/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?

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):

Code:

---

#!/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

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 $. Like, what is the meaning of "$?" ..?

Quote:

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Originally Posted by 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):

Code:

---

#!/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

---

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

Quote:

---

Originally Posted by 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

---

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:

Code:

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man bash | less +/" +  Parameter Expansion"

---

$ is the expansion character. It's a special character or metacharacter used in substitutions: parameter and variable expansion, command substitution and arithmetic expansion.


? and $ (yep, whatever includes the $ sign too :) ) are special 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.

Quote:

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Originally Posted by sisco311

That link contains inaccurate informations.

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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.