Question

2

How to check file size and then echo?

rated 0 times [ 2] [ 0] / answers: 1 / hits: 636 / 3 Years ago, thu, august 19, 2021, 6:07:03

Currently I'm trying to check file size from reading a directory

read directory



dirsize=$(du -sh $directory)



if [[ "$dirsize" -ge 2MB ]]; then

  echo 'High'



elif [[ "$dirsize" -le 2MB ]]; then

  echo 'Low'



fi

The problem is when I check file size, the output will be like

4.0K ./......./...
32M ./......./...

So, when I write my code it doesn't work and I don't know how to fix.
For example,

if [[ "$dirsize" -ge 2MB ]]; then

   echo 'High'

From code above, my point is to echo 'High' if file is greater than 2MB.
Could you give advice please.

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wherejuic

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answered 3 Years ago urnaetted · Accepted Answer

du returns the size and the file/directory name. As you only need the size you just read the first part. And you need the bytes, not the iec size.

You can use numfmt to convert iec numbers to bytes (see man numfmt).

Numbers can be compared by bash with number compare operators (see man test).

# use only the first part of the output from "du" and use size in bytes

read -r size _ <<<"$( du -sb directory )"

# convert 2M to bytes

size2m="$( numfmt --from=iec 2M )"



# now compare the pure numbers

if [[ ${size} -ge ${size2m} ]]; then

    echo "High"

fi



if [[ ${size} -le ${size2m} ]]; then

    echo "Low"

fi