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up vote 1 down vote favorite

Data

1
\begin{document}
3

Code 

#!/bin/bash

function getStart {
        local START="$(awk '/begin\{document\}/{ print NR; exit }' data.tex)"
        echo $START
}

START2=$(getStart)
echo $START2

which returns 2 but I want 3. I change unsuccessfully the end by this answer about How can I add numbers in a bash script:

START2=$((getStart+1))

How can you increment a local variable in Bash script?

share|improve this question
    
I'm getting 2, not 1, from the code. – choroba Sep 11 '15 at 13:03
    
Sorry my mistake! – Masi Sep 11 '15 at 13:13
1  
OFF: why awk? sed -n '/begin{document}/{=;q}' data.text much shorter… – Costas Sep 11 '15 at 13:46
    
@Costas Yes, you are right! I have had today a bad day in thinking too complicated. Thinking now the thing here for open intervals: unix.stackexchange.com/q/229060/16920 Can you explain }/{=;q} this in an answer/comment, please? – Masi Sep 11 '15 at 13:47
up vote 7 down vote accepted

I'm getting 2 from your code. Nevertheless, you can use the same technique for any variable or number:

local start=1
(( start++ ))

or

(( ++start ))

or

(( start += 1 ))

or

(( start = start + 1 ))

or just

local start=1
echo $(( start + 1 ))

etc.

share|improve this answer
    
This may also help: askubuntu.com/questions/385528/… – Bruno Bieri Oct 12 at 6:30
up vote 1 down vote

Try:

START2=$(( `getStart` + 1 ));

The $(( )) tells bash that it is to perform an arithmetic operation, while the backticks tells bash to evaluate the containing expression, be it an user-defined function or a call to an external program, and return the contents of stdout.

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