floating-point operations with bash

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无人共我
无人共我 2021-02-09 05:24

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 f

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  • 2021-02-09 05:46

    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
    
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  • 2021-02-09 06:04

    Use bc -l

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

    OR awk:

    awk 'BEGIN{printf "%.2f\n", (620/100)}'
    6.20
    
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  • 2021-02-09 06:07

    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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