Extracting text for a line using bash
 

I am completely new to linux and bash.

Terminal shows a number of lines as the output of the 'xset q'.
I would like to extract the word after the first word 'timeout:' on a line from the result of 'xset q'.

I guess using 'if' could be a starting point. But this needs a lot more processing.
Could someone help?

Look into grep and awk

As a hint. Not exactly sure how you define word but the awk command above splits the results by space and {print $2} outputs the second field. From your other thread if you post the batch file you wish to convert we can provide better help.

That's great and simple.Exactly what I was looking for.

Next, the if statement syntax looks complicated.
What I would like to do is:
if (xset q | grep timeout | awk -F ' ' '{print $2}') is not zero (0), then execute 'xset s off'
This should be easy if the syntax is right:

if [[xset q | grep timeout | awk -F ' ' '{print $2}' != "0"]]
then xset s off
fi

Saved above in a file trial.sh
Tried to run it in terminal with the following error:
/mnt/Data/Temp/trial1.sh: line 3: [[xset: command not found
awk: fatal: cannot open file `=' for reading: No such file or directory

Edit: The batch file from the other thread is too complicated to start with.

bash has some unique syntax requirements, spaces in certain locations do matter as well as where you place keywords. You can check your syntax by copying your script to https://www.shellcheck.net/
This isn't the only way...

awk is more than capable; the pipe via grep is superfluous - and somewhat surprising it was suggested.

Strictly speaking bash is also capable without the external calls at all.

Ahhh - I see michaelk got in first with suggested code.

I know grep is superfluous, it just happened...
something like

I entered your suggestion into ShellCheck and provided the following:

$ shellcheck myscript
No issues detected!

I entered it into my .sh file and worked.
I see what you mean by the correct placing of spaces.

I wouldn't mind trying the simpler version.

Edit: Didn't see your your last post michaelk.
Tried it and also works.

syg00 is much better at scripting then I.

That looks great.
What I don't follow is the "$1 ~ /timeout/"?
I cant see that in the ss64.com about awk.

Basically awk searches the first column ($1) for the value timeout based on using the space as a column separator (-F ' '). Given the desired output of "timeout: 0 cycle: 0" the variables are assigned as $0 being the entire string, $1 as timeout:, $2 as 0, $3 cycle: and $4 as 0.

Thank you. I thought that was the case but ss64.com shows an example:

So an AWK program to retrieve Audrey's phone number is:
'$1 == "Audrey" {print $2}'

So, I didn't understand and still don't $1 ~ /timeout/ with the "~" and forward slashes.

I did try the SS64.com example and doesn't work properly.

For an exact match
The ~ is a regular expression that basically means match column that starts with timeout

Now, that does work.
The change I see is the addition of the ":" at the end of the timeout. Maybe because the contents of $1 is actually timeout:.
This is what I showed originally. I must admit I didn't understand dropping it for the /timeout/ version.
I still don't know the meaning of the "~" and the forward slashes. Maybe the "~" means contains timeout - don't know but it works.

see man awk about how does it work or here: https://www.gnu.org/software/gawk/ma...exp-Usage.html