Shell: Robust way to remove trailing decimal point and zero in case there is a zero
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
add a comment |
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
2
With bash:echo "${foo//.0/}"
?
– Cyrus
Jan 1 at 10:52
@Cyrus8.04
ends up being84
– tzippy
Jan 8 at 14:14
add a comment |
In a bash script I have a (two digit) number with decimal point and one digit following.
In case there's a zero after the decimal separator, I want to trim the separator and the zero.
For
foo=26.5
..there should be no trimming. But in this case:
foo=26.0
My desired output would be:
foo=26
Is tr
the right choice for it?
I tried it with the --delete .0
option but it simply deletes all of both character's appearences.
Thank you!
bash shell trim tr
In a bash script I have a (two digit) number with decimal point and one digit following.
In case there's a zero after the decimal separator, I want to trim the separator and the zero.
For
foo=26.5
..there should be no trimming. But in this case:
foo=26.0
My desired output would be:
foo=26
Is tr
the right choice for it?
I tried it with the --delete .0
option but it simply deletes all of both character's appearences.
Thank you!
bash shell trim tr
bash shell trim tr
edited Jan 1 at 10:53
tzippy
asked Jan 1 at 10:44
tzippytzippy
2,6991757105
2,6991757105
2
With bash:echo "${foo//.0/}"
?
– Cyrus
Jan 1 at 10:52
@Cyrus8.04
ends up being84
– tzippy
Jan 8 at 14:14
add a comment |
2
With bash:echo "${foo//.0/}"
?
– Cyrus
Jan 1 at 10:52
@Cyrus8.04
ends up being84
– tzippy
Jan 8 at 14:14
2
2
With bash:
echo "${foo//.0/}"
?– Cyrus
Jan 1 at 10:52
With bash:
echo "${foo//.0/}"
?– Cyrus
Jan 1 at 10:52
@Cyrus
8.04
ends up being 84
– tzippy
Jan 8 at 14:14
@Cyrus
8.04
ends up being 84
– tzippy
Jan 8 at 14:14
add a comment |
1 Answer
1
active
oldest
votes
Not only for bash
:
foo=$(echo "$foo" | sed 's/.0$//')
why not simply foo="${foo%.*}"
– ctac_
Jan 1 at 16:17
@ctac_: This probably only works forbash
and removes everything after the dot. I suggest to replace*
with0
.
– Cyrus
Jan 1 at 18:04
I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?
– ctac_
Jan 1 at 18:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Not only for bash
:
foo=$(echo "$foo" | sed 's/.0$//')
why not simply foo="${foo%.*}"
– ctac_
Jan 1 at 16:17
@ctac_: This probably only works forbash
and removes everything after the dot. I suggest to replace*
with0
.
– Cyrus
Jan 1 at 18:04
I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?
– ctac_
Jan 1 at 18:17
add a comment |
Not only for bash
:
foo=$(echo "$foo" | sed 's/.0$//')
why not simply foo="${foo%.*}"
– ctac_
Jan 1 at 16:17
@ctac_: This probably only works forbash
and removes everything after the dot. I suggest to replace*
with0
.
– Cyrus
Jan 1 at 18:04
I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?
– ctac_
Jan 1 at 18:17
add a comment |
Not only for bash
:
foo=$(echo "$foo" | sed 's/.0$//')
Not only for bash
:
foo=$(echo "$foo" | sed 's/.0$//')
answered Jan 1 at 11:27


CyrusCyrus
46.4k43880
46.4k43880
why not simply foo="${foo%.*}"
– ctac_
Jan 1 at 16:17
@ctac_: This probably only works forbash
and removes everything after the dot. I suggest to replace*
with0
.
– Cyrus
Jan 1 at 18:04
I think it's ok with all POSIX shell. It seems to me the OP don't tell all. Starting from 26.5, he want to get 26 ?
– ctac_
Jan 1 at 18:17
add a comment |
why not simply foo="${foo%.*}"
– ctac_
Jan 1 at 16:17
@ctac_: This probably only works forbash
and removes everything after the dot. I suggest to replace*
with0
.
– 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
add a comment |
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2
With bash:
echo "${foo//.0/}"
?– Cyrus
Jan 1 at 10:52
@Cyrus
8.04
ends up being84
– tzippy
Jan 8 at 14:14