Question

21

How to pass string variable as parameter to awk

rated 0 times [ 21] [ 0] / answers: 1 / hits: 69638 / 3 Years ago, tue, july 6, 2021, 3:23:09

I have following command:

usb_soundcard_sink=$(pactl list short sinks | grep "alsa_output" | awk '{ print $2 }' | tail -n1)

output of this command is:

alsa_output.pci-0000_00_1b.0.analog-stereo

And this is another command that find index number:

var=$(pactl list short sinks | awk '$2=="alsa_output.pci-0000_00_1b.0.analog-stereo" {print $1}')

output:

But I want pass the "usb_soundcard_sink" variable instead of hard coded value i.e. "alsa_output.pci-0000_00_1b.0.analog-stereo" in above command. b'coz value of "usb_soundcard_sink" variable may change dynamically.

Also I tried the following:

var=$(pactl list short sinks | awk '$2=="$usb_soundcard_sink" {print $1}')

But it is not working

so how can I pass value of "usb_soundcard_sink" variable to the above command

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termetalli

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4 Years ago

answered 3 Years ago taigysel · Accepted Answer

To pass a inside into your awk code, you have to use the -v syntax like this:

$ awk -v myvar="3" 'BEGIN{print myvar}'

#     __________/              ___/          

3

In your specific case, just use:

pactl list short sinks | awk -v myvar="$usb_soundcard_sink" '$2==myvar {print $1}'

#                               ___/                            ___/

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