How can I convert a string to an integer within Bash?

Question

I am attempting to create a cronjob that will run depending on the output of disk usage for a specific folder. In my test environment, I have a folder called "Test" that is 4.0 Kilobytes. I am querying its size with:

du -csh /home/<username>/Test | grep total | cut -f1 | tr -d 'K'

The output is assigned to a variable called DISK_SPACE and leaves me with 4.0. I cannot figure out how to convert 4.0 to an integer in order to satisfy the following:

if [ $DISK_SPACE -gt 3.0 ]
then
rm -rf /home/<username>/Test/*.*
else
echo $DISK_SPACE "is less than 3.0K and as a result the contents of the folder will NOT be deleted."

My full bash file looks like the below:

#!/bin/bash
DISK_SPACE=$(du -csh /home/<username>/Test | grep total | cut -f1 | tr -d 'K')
echo $DISK_SPACE
if [ $DISK_SPACE -gt 3.0 ]
then
rm -rf /home/<username>Test/*.*
else
echo $DISK_SPACE "is less than 3.0K and as a result the contents of the folder will NOT be deleted."
fi

The error I receive after running this is:

4.0: integer expression expected

Your title suggests a more general question than just one involving disk utilization computation and comparison. — JRFerguson, Mar 19 '19 at 19:17

score 4 · Accepted Answer · answered Mar 19 '19 at 18:41

4

du -h prints numbers for human consumption. A script shouldn't use -h. Try -k or -b instead to get easy-to-parse integers:

-k     like --block-size=1K

-b, --bytes
       equivalent to '--apparent-size --block-size=1'

answered Mar 19 '19 at 18:41

John Kugelman

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JRFerguson · Answer 2 · 2019-03-19T18:46:22.367

0

Use printf to round your real number to an integer.

printf "%1.0f\n" 4.0
4

printf "%1.0f\n" 4.6
5

See the printf manpages.

For you example you can do:

DISK_SPACE=4.0
DISK_SPACE=$(printf "%1.0f" ${DISK_SPACE})
echo ${DISK_SPACE}
4

edited Mar 19 '19 at 18:46

answered Mar 19 '19 at 18:37

JRFerguson

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1

Under [tag:bash] use `printf -v DISK_SPACE %.0f $DISKSPACE`. This will avoid one fork and use builtin `printf -v` to set variable. – F. Hauri - Give Up GitHub Mar 19 '19 at 21:10

tripleee · Answer 3 · 2019-03-19T19:12:24.163

0

Bash has no support for floating point arithmetic. If you just care whether the first digit is bigger than 3, you can simply trim any decimals.

if [ "${disk_space%.*}" -gt 3 ]; then ...

If you need proper float comparison, perhaps it's simpler to actually use a tool which supports floating-point arithmetic.

Notice also that I converted your variable to lower case. You should not use all-uppercase names for your private variables.

edited Mar 19 '19 at 19:12

answered Mar 19 '19 at 18:47

tripleee

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Your second example is missing a closing `"` on the variable expansion. And it's also broken in that it dies with an error for `disk_space=3.0`. – Mike Holt Mar 19 '19 at 19:07
Thanks, removed that portion, which was kind of silly anyway. – tripleee Mar 19 '19 at 19:14
Yeah, it was probably overkill. But it actually would have worked if you'd replaced the `-eq` with `=` (i.e., just doing regular string comparison instead of a numeric "equals" test). In that case, things like `"3.0000"`, etc would just stay as they were and therefore fail the string comparison as intended, while things like `"3.00195"` would be chopped down to `"3"`. – Mike Holt Mar 19 '19 at 19:17

score 0 · Answer 4 · edited Mar 20 '19 at 18:15

0

In addition to John Kugelman's answer I also changed the following line:

Original:

if [ $DISK_SPACE -gt 3.0 ]

New:

if [ $DISK_SPACE -gt 3072 ]

This allowed John's recommended change to work even better with the script:

du -csb /home/<username>/Test | grep total | cut -f1

edited Mar 20 '19 at 18:15

John Kugelman

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answered Mar 19 '19 at 19:28

ARedPaladin

23
4

How can I convert a string to an integer within Bash?

4 Answers4