Domanda

how can I transform the string "620/100" into "6.2" in a bash script

The context of my question is about image processing. EXIF data are coding the focal length in fractional format, while I need the corresponding decimal string.

Thanks for helping, Olivier

È stato utile?

Soluzione

Use bc -l

bc -l <<< "scale=2; 620/100"
6.20

OR awk:

awk 'BEGIN{printf "%.2f\n", (620/100)}'
6.20

Altri suggerimenti

bash doesn't support floating point.

You could use bc:

$ echo "50/10" | bc -l
5.00000000000000000000
$ echo "scale=1; 50/10" | bc -l
5.0

Thank-you for the answers. bc was what I needed.

I dont know if posting the result is of any use. Anyway, this the final piece of code for exteracting the focal length of a photo and print it in decimal format. It is supposed to work for all cameras. Tested on 4 cameras of 3 different brands.

F="your_image.JPG"
EXIF=$(exiv2 -p v "$F")
FocalFractional=$( echo "$EXIF" | grep -E '[^ ]*  *Photo *FocalLength '| grep -iohE "[^ ]* *$" )
Formula="scale=2; "$FocalFractional
FocalDecimal=$( bc -l <<< "$Formula" )
echo $ FocalDecimal
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