Shell: Robust way to remove trailing decimal point and zero in case there is a zero












0















In a bash script I have a (two digit) number with decimal point and one digit following.



In case there's a zero after the decimal separator, I want to trim the separator and the zero.



For



foo=26.5


..there should be no trimming. But in this case:



foo=26.0 


My desired output would be:



foo=26


Is tr the right choice for it?



I tried it with the --delete .0 option but it simply deletes all of both character's appearences.



Thank you!










share|improve this question




















  • 2





    With bash: echo "${foo//.0/}"?

    – Cyrus
    Jan 1 at 10:52













  • @Cyrus 8.04 ends up being 84

    – tzippy
    Jan 8 at 14:14
















0















In a bash script I have a (two digit) number with decimal point and one digit following.



In case there's a zero after the decimal separator, I want to trim the separator and the zero.



For



foo=26.5


..there should be no trimming. But in this case:



foo=26.0 


My desired output would be:



foo=26


Is tr the right choice for it?



I tried it with the --delete .0 option but it simply deletes all of both character's appearences.



Thank you!










share|improve this question




















  • 2





    With bash: echo "${foo//.0/}"?

    – Cyrus
    Jan 1 at 10:52













  • @Cyrus 8.04 ends up being 84

    – tzippy
    Jan 8 at 14:14














0












0








0








In a bash script I have a (two digit) number with decimal point and one digit following.



In case there's a zero after the decimal separator, I want to trim the separator and the zero.



For



foo=26.5


..there should be no trimming. But in this case:



foo=26.0 


My desired output would be:



foo=26


Is tr the right choice for it?



I tried it with the --delete .0 option but it simply deletes all of both character's appearences.



Thank you!










share|improve this question
















In a bash script I have a (two digit) number with decimal point and one digit following.



In case there's a zero after the decimal separator, I want to trim the separator and the zero.



For



foo=26.5


..there should be no trimming. But in this case:



foo=26.0 


My desired output would be:



foo=26


Is tr the right choice for it?



I tried it with the --delete .0 option but it simply deletes all of both character's appearences.



Thank you!







bash shell trim tr






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 1 at 10:53







tzippy

















asked Jan 1 at 10:44









tzippytzippy

2,6991757105




2,6991757105








  • 2





    With bash: echo "${foo//.0/}"?

    – Cyrus
    Jan 1 at 10:52













  • @Cyrus 8.04 ends up being 84

    – tzippy
    Jan 8 at 14:14














  • 2





    With bash: echo "${foo//.0/}"?

    – Cyrus
    Jan 1 at 10:52













  • @Cyrus 8.04 ends up being 84

    – tzippy
    Jan 8 at 14:14








2




2





With bash: echo "${foo//.0/}"?

– Cyrus
Jan 1 at 10:52







With bash: echo "${foo//.0/}"?

– Cyrus
Jan 1 at 10:52















@Cyrus 8.04 ends up being 84

– tzippy
Jan 8 at 14:14





@Cyrus 8.04 ends up being 84

– tzippy
Jan 8 at 14:14












1 Answer
1






active

oldest

votes


















1














Not only for bash:



foo=$(echo "$foo" | sed 's/.0$//')





share|improve this answer
























  • why not simply foo="${foo%.*}"

    – ctac_
    Jan 1 at 16:17











  • @ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

    – Cyrus
    Jan 1 at 18:04













  • I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

    – ctac_
    Jan 1 at 18:17











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Not only for bash:



foo=$(echo "$foo" | sed 's/.0$//')





share|improve this answer
























  • why not simply foo="${foo%.*}"

    – ctac_
    Jan 1 at 16:17











  • @ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

    – Cyrus
    Jan 1 at 18:04













  • I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

    – ctac_
    Jan 1 at 18:17
















1














Not only for bash:



foo=$(echo "$foo" | sed 's/.0$//')





share|improve this answer
























  • why not simply foo="${foo%.*}"

    – ctac_
    Jan 1 at 16:17











  • @ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

    – Cyrus
    Jan 1 at 18:04













  • I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

    – ctac_
    Jan 1 at 18:17














1












1








1







Not only for bash:



foo=$(echo "$foo" | sed 's/.0$//')





share|improve this answer













Not only for bash:



foo=$(echo "$foo" | sed 's/.0$//')






share|improve this answer












share|improve this answer



share|improve this answer










answered Jan 1 at 11:27









CyrusCyrus

46.4k43880




46.4k43880













  • why not simply foo="${foo%.*}"

    – ctac_
    Jan 1 at 16:17











  • @ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

    – Cyrus
    Jan 1 at 18:04













  • I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

    – ctac_
    Jan 1 at 18:17



















  • why not simply foo="${foo%.*}"

    – ctac_
    Jan 1 at 16:17











  • @ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

    – Cyrus
    Jan 1 at 18:04













  • I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

    – ctac_
    Jan 1 at 18:17

















why not simply foo="${foo%.*}"

– ctac_
Jan 1 at 16:17





why not simply foo="${foo%.*}"

– ctac_
Jan 1 at 16:17













@ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

– Cyrus
Jan 1 at 18:04







@ctac_: This probably only works for bash and removes everything after the dot. I suggest to replace * with 0.

– Cyrus
Jan 1 at 18:04















I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

– ctac_
Jan 1 at 18:17





I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?

– ctac_
Jan 1 at 18:17




















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