CODEWITHSUNDEEP
shellsh
Kevin R.
Kevin R.

Reputation: 154

Use a shell output value in shell script

If I try following: 

varReturn=$(ls)
echo $varReturn

it shows me the correct output of the listed elements in the directory.

But if I try this one: 

varReturn=$(/opt/vc/bin/tvservice -n)
echo $varReturn

it doesn't show me the expected output :/

My goal is to check if an HDMI Port is connected or not. It' very curious for me, why it works only for some commands.

I'm looking forward to getting some help here. I didn't figure out, what the problem is.

EDIT:

Now I've found another way and tried following: 

varReturn=`tvservice -s`
echo $varReturn

this shows me the correct output:

enter image description here

But if I use another command, like this one:

varReturn=`tvservice -n`
echo $varReturn

It shows me no output at echo, but the output from the var (confusing). enter image description here

It still shows me the output if I use following code:

varReturn=`tvservice -n`
#echo $varReturn

The output is shown without the blank space.

Upvotes: 1

Views: 102

Answers (2)

Abhijit Pritam Dutta
Abhijit Pritam Dutta

Reputation: 5591

When you execute a shell command like varReturn=$(/opt/vc/bin/tvservice -n) it will store the output to the variable only when the command executed successfully, else it will not hold any information because error/unsuccessful message will be redirected to standard error. Hence you have to redirected it to standard output like below:-

varReturn=$(/opt/vc/bin/tvservice -n 2>&1)

Now in both successful and unsuccessful execution case output will store in variable varReturn.

Upvotes: 1

Tom Fenech
Tom Fenech

Reputation: 74615

There is at least one problem with this code:

varReturn=$(/opt/vc/bin/tvservice -n)
echo $varReturn
#    ^ missing double quotes around this variable

Adding those quotes will ensure that the variable is passed as a single argument to echo. Otherwise, echo sees a list of arguments and outputs each one, separated by a space.

The next possible issue is that the command is outputting to standard error, rather than standard output, so it won't be captured by $() or the old-fashioned equivalent ` `. To correct this, try:

output=$(/opt/vc/bin/tvservice -n 2>&1)
#                                 ^ redirect standard error to standard output
echo "$output"

Upvotes: 1

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