CODEWITHSUNDEEP
bashsed
mr. cooper
mr. cooper

Reputation: 880

Bash math then pass output to sed

I'd like to use the index of a for loop in bash to calculate a number then pass that number to sed:

for i in {1989..2009..1}
do
let year=i-1
echo $year
find /cygdrive/d/snowmodel/$i/snowmodel.par -type f -exec \
sed -i 's/iyear_init = 1989/iyear_init = '$year'/g' {} +
done

So I want to replace the line iyear_init = 1989 with the line iyear_init = (the value of i-1, which should be the variable "year").

the echo $year command returns the correct value, but it seems that when it gets passed to sed it reverts back to treating year like a string.

Thanks for any help.

Upvotes: 1

Views: 219

Answers (1)

mr. cooper
mr. cooper

Reputation: 880

Ok found answer after stackoverflow suggested Related questions (I guess my search queries weren't worded very well).

The problem was I needed to replace the single quotes with double quotes because variables inside single quotes don't get replaced in sed. The following worked:

for i in {1989..2009..1}
do
let year=i-1
echo $year
find /cygdrive/d/snowmodel_cig/cig_79/bias_runs/$i/snowmodel.par -type f -exec \
sed -i "s/iyear_init = i-1/iyear_init = $year/g" {} +
done

Upvotes: 1

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