Variable in Date command
Clash Royale CLAN TAG#URR8PPP
I want to use a variable inside date
command . GNU date is not supportable in my system.
I use Sunos and when I use date
with -d
option it is shwoing -d
as bad substitution.
#!/usr/bin/ksh
RELEASE_DATE="28 OCT 2018"
RELEASE_DATE_MINUS_2=`date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'`
echo "Release is $RELEASE_DATE"
echo "release_date-2 is $RELEASE_DATE_MINUS_2
I have tried the above code snippet.
I should pass the actual release_date
which i get it form another .cnf
file
like 28OCT2018
with this input I should get two days before that date like 2018-10-26
.
shell scripting solaris
|
show 3 more comments
I want to use a variable inside date
command . GNU date is not supportable in my system.
I use Sunos and when I use date
with -d
option it is shwoing -d
as bad substitution.
#!/usr/bin/ksh
RELEASE_DATE="28 OCT 2018"
RELEASE_DATE_MINUS_2=`date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'`
echo "Release is $RELEASE_DATE"
echo "release_date-2 is $RELEASE_DATE_MINUS_2
I have tried the above code snippet.
I should pass the actual release_date
which i get it form another .cnf
file
like 28OCT2018
with this input I should get two days before that date like 2018-10-26
.
shell scripting solaris
1
What language isrel_date
a variable in?
– Jesse_b
Jan 7 at 13:56
Will your input date always be in the same format? Also will it pad day numbers less than 10? ie2019JAN07
or2019JAN7
?
– Jesse_b
Jan 7 at 14:03
What OS and/or Distribution is this running on? (Do you have GNUdate
?)
– roaima
Jan 7 at 14:08
3
"I have used various date command features" care to share which ones you've tried, and how? Or do we need to guess?
– roaima
Jan 7 at 14:08
2
"I am getting the error in stating -d option is not a valid argument." What command did you use, and what was the exact error message? (Don't approximate.)
– roaima
Jan 7 at 14:09
|
show 3 more comments
I want to use a variable inside date
command . GNU date is not supportable in my system.
I use Sunos and when I use date
with -d
option it is shwoing -d
as bad substitution.
#!/usr/bin/ksh
RELEASE_DATE="28 OCT 2018"
RELEASE_DATE_MINUS_2=`date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'`
echo "Release is $RELEASE_DATE"
echo "release_date-2 is $RELEASE_DATE_MINUS_2
I have tried the above code snippet.
I should pass the actual release_date
which i get it form another .cnf
file
like 28OCT2018
with this input I should get two days before that date like 2018-10-26
.
shell scripting solaris
I want to use a variable inside date
command . GNU date is not supportable in my system.
I use Sunos and when I use date
with -d
option it is shwoing -d
as bad substitution.
#!/usr/bin/ksh
RELEASE_DATE="28 OCT 2018"
RELEASE_DATE_MINUS_2=`date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'`
echo "Release is $RELEASE_DATE"
echo "release_date-2 is $RELEASE_DATE_MINUS_2
I have tried the above code snippet.
I should pass the actual release_date
which i get it form another .cnf
file
like 28OCT2018
with this input I should get two days before that date like 2018-10-26
.
shell scripting solaris
shell scripting solaris
edited Jan 11 at 11:45
chaos
35.3k773117
35.3k773117
asked Jan 7 at 13:54
SandraSandra
112
112
1
What language isrel_date
a variable in?
– Jesse_b
Jan 7 at 13:56
Will your input date always be in the same format? Also will it pad day numbers less than 10? ie2019JAN07
or2019JAN7
?
– Jesse_b
Jan 7 at 14:03
What OS and/or Distribution is this running on? (Do you have GNUdate
?)
– roaima
Jan 7 at 14:08
3
"I have used various date command features" care to share which ones you've tried, and how? Or do we need to guess?
– roaima
Jan 7 at 14:08
2
"I am getting the error in stating -d option is not a valid argument." What command did you use, and what was the exact error message? (Don't approximate.)
– roaima
Jan 7 at 14:09
|
show 3 more comments
1
What language isrel_date
a variable in?
– Jesse_b
Jan 7 at 13:56
Will your input date always be in the same format? Also will it pad day numbers less than 10? ie2019JAN07
or2019JAN7
?
– Jesse_b
Jan 7 at 14:03
What OS and/or Distribution is this running on? (Do you have GNUdate
?)
– roaima
Jan 7 at 14:08
3
"I have used various date command features" care to share which ones you've tried, and how? Or do we need to guess?
– roaima
Jan 7 at 14:08
2
"I am getting the error in stating -d option is not a valid argument." What command did you use, and what was the exact error message? (Don't approximate.)
– roaima
Jan 7 at 14:09
1
1
What language is
rel_date
a variable in?– Jesse_b
Jan 7 at 13:56
What language is
rel_date
a variable in?– Jesse_b
Jan 7 at 13:56
Will your input date always be in the same format? Also will it pad day numbers less than 10? ie
2019JAN07
or 2019JAN7
?– Jesse_b
Jan 7 at 14:03
Will your input date always be in the same format? Also will it pad day numbers less than 10? ie
2019JAN07
or 2019JAN7
?– Jesse_b
Jan 7 at 14:03
What OS and/or Distribution is this running on? (Do you have GNU
date
?)– roaima
Jan 7 at 14:08
What OS and/or Distribution is this running on? (Do you have GNU
date
?)– roaima
Jan 7 at 14:08
3
3
"I have used various date command features" care to share which ones you've tried, and how? Or do we need to guess?
– roaima
Jan 7 at 14:08
"I have used various date command features" care to share which ones you've tried, and how? Or do we need to guess?
– roaima
Jan 7 at 14:08
2
2
"I am getting the error in stating -d option is not a valid argument." What command did you use, and what was the exact error message? (Don't approximate.)
– roaima
Jan 7 at 14:09
"I am getting the error in stating -d option is not a valid argument." What command did you use, and what was the exact error message? (Don't approximate.)
– roaima
Jan 7 at 14:09
|
show 3 more comments
4 Answers
4
active
oldest
votes
With the zsh
shell:
zmodload zsh/datetime
rel_date=2018OCT28
strftime -rs d %Y%b%d.%H $rel_date.12 &&
strftime -s two_days_before %Y-%m-%d $((d - 86400*2)) || exit
echo $two_days_before
add a comment |
You tagged this with shell so I will assume the variable is supposed to be in shell but FYI you cannot have spaces around the =
for a shell variable.
GNU date
will not accept your input date in that format but will accept it like 28 OCT 2018
so using shell parameter expansion we can rearrange the date (as long as it will always be in the same format):
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4"
Now in order to subtract 2 days we simply need to add:
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days"
And to have it output in your requested format
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days" '+%Y-%m-%d'
We don't (yet) know for sure the OP is using GNUdate
.
– roaima
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"
– Sandra
Jan 8 at 14:19
|
show 3 more comments
Perl can be useful for this:
perl -MTime::Piece -slE 'say +(Time::Piece->strptime($date, "%Y%b%d") - 2 * 86_400)->ymd' -- -date="$rel_date"
You specify the format of the incoming date variable to parse it into a time object, do the date arithmetic, then output into the desired YYYYY-mm-dd format.
Documented here: https://perldoc.perl.org/Time/Piece.html
add a comment |
It could be useful to know which ksh88,93 are you using.
According to ksh93(1) you can find something about date:
printf format [ arg ... ]
%(date-format)T
--practical example
: printf "%(%d.%m,%Y)Tn"
11.01,2019
: printf "%(%d.%m.%Y)Tn" "2 days ago"
09.01.2019
Let's hope it helps you.
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "106"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f492998%2fvariable-in-date-command%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
With the zsh
shell:
zmodload zsh/datetime
rel_date=2018OCT28
strftime -rs d %Y%b%d.%H $rel_date.12 &&
strftime -s two_days_before %Y-%m-%d $((d - 86400*2)) || exit
echo $two_days_before
add a comment |
With the zsh
shell:
zmodload zsh/datetime
rel_date=2018OCT28
strftime -rs d %Y%b%d.%H $rel_date.12 &&
strftime -s two_days_before %Y-%m-%d $((d - 86400*2)) || exit
echo $two_days_before
add a comment |
With the zsh
shell:
zmodload zsh/datetime
rel_date=2018OCT28
strftime -rs d %Y%b%d.%H $rel_date.12 &&
strftime -s two_days_before %Y-%m-%d $((d - 86400*2)) || exit
echo $two_days_before
With the zsh
shell:
zmodload zsh/datetime
rel_date=2018OCT28
strftime -rs d %Y%b%d.%H $rel_date.12 &&
strftime -s two_days_before %Y-%m-%d $((d - 86400*2)) || exit
echo $two_days_before
edited Jan 7 at 14:22
answered Jan 7 at 14:17
Stéphane ChazelasStéphane Chazelas
301k55566918
301k55566918
add a comment |
add a comment |
You tagged this with shell so I will assume the variable is supposed to be in shell but FYI you cannot have spaces around the =
for a shell variable.
GNU date
will not accept your input date in that format but will accept it like 28 OCT 2018
so using shell parameter expansion we can rearrange the date (as long as it will always be in the same format):
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4"
Now in order to subtract 2 days we simply need to add:
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days"
And to have it output in your requested format
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days" '+%Y-%m-%d'
We don't (yet) know for sure the OP is using GNUdate
.
– roaima
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"
– Sandra
Jan 8 at 14:19
|
show 3 more comments
You tagged this with shell so I will assume the variable is supposed to be in shell but FYI you cannot have spaces around the =
for a shell variable.
GNU date
will not accept your input date in that format but will accept it like 28 OCT 2018
so using shell parameter expansion we can rearrange the date (as long as it will always be in the same format):
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4"
Now in order to subtract 2 days we simply need to add:
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days"
And to have it output in your requested format
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days" '+%Y-%m-%d'
We don't (yet) know for sure the OP is using GNUdate
.
– roaima
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"
– Sandra
Jan 8 at 14:19
|
show 3 more comments
You tagged this with shell so I will assume the variable is supposed to be in shell but FYI you cannot have spaces around the =
for a shell variable.
GNU date
will not accept your input date in that format but will accept it like 28 OCT 2018
so using shell parameter expansion we can rearrange the date (as long as it will always be in the same format):
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4"
Now in order to subtract 2 days we simply need to add:
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days"
And to have it output in your requested format
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days" '+%Y-%m-%d'
You tagged this with shell so I will assume the variable is supposed to be in shell but FYI you cannot have spaces around the =
for a shell variable.
GNU date
will not accept your input date in that format but will accept it like 28 OCT 2018
so using shell parameter expansion we can rearrange the date (as long as it will always be in the same format):
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4"
Now in order to subtract 2 days we simply need to add:
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days"
And to have it output in your requested format
date -d "$rel_date: -2 $rel_date:4:3 $rel_date:0:4 - 2 days" '+%Y-%m-%d'
edited Jan 7 at 14:18
answered Jan 7 at 14:11
Jesse_bJesse_b
12.1k23064
12.1k23064
We don't (yet) know for sure the OP is using GNUdate
.
– roaima
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"
– Sandra
Jan 8 at 14:19
|
show 3 more comments
We don't (yet) know for sure the OP is using GNUdate
.
– roaima
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"
– Sandra
Jan 8 at 14:19
We don't (yet) know for sure the OP is using GNU
date
.– roaima
Jan 7 at 14:12
We don't (yet) know for sure the OP is using GNU
date
.– roaima
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
We don't but I have provided a GNU date solution.
– Jesse_b
Jan 7 at 14:12
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
Hi Jess, Thanks for your help. But even this seems to be not working .
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
It just tells the same bad substitution error . Below is my code snipet.
– Sandra
Jan 8 at 14:18
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=
date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"– Sandra
Jan 8 at 14:19
#!/usr/bin/ksh RELEASE_DATE="28 OCT 2018" RELEASE_DATE_MINUS_2=
date -"$RELEASE_DATE: -2 $RELEASE_DATE:4:3 $RELEASE_DATE:0:4 - 2 days" '+%Y-%m-%d'
echo "Release is $RELEASE_DATE" echo "releasE_date-2 is $RELEASE_DATE_MINUS_2"– Sandra
Jan 8 at 14:19
|
show 3 more comments
Perl can be useful for this:
perl -MTime::Piece -slE 'say +(Time::Piece->strptime($date, "%Y%b%d") - 2 * 86_400)->ymd' -- -date="$rel_date"
You specify the format of the incoming date variable to parse it into a time object, do the date arithmetic, then output into the desired YYYYY-mm-dd format.
Documented here: https://perldoc.perl.org/Time/Piece.html
add a comment |
Perl can be useful for this:
perl -MTime::Piece -slE 'say +(Time::Piece->strptime($date, "%Y%b%d") - 2 * 86_400)->ymd' -- -date="$rel_date"
You specify the format of the incoming date variable to parse it into a time object, do the date arithmetic, then output into the desired YYYYY-mm-dd format.
Documented here: https://perldoc.perl.org/Time/Piece.html
add a comment |
Perl can be useful for this:
perl -MTime::Piece -slE 'say +(Time::Piece->strptime($date, "%Y%b%d") - 2 * 86_400)->ymd' -- -date="$rel_date"
You specify the format of the incoming date variable to parse it into a time object, do the date arithmetic, then output into the desired YYYYY-mm-dd format.
Documented here: https://perldoc.perl.org/Time/Piece.html
Perl can be useful for this:
perl -MTime::Piece -slE 'say +(Time::Piece->strptime($date, "%Y%b%d") - 2 * 86_400)->ymd' -- -date="$rel_date"
You specify the format of the incoming date variable to parse it into a time object, do the date arithmetic, then output into the desired YYYYY-mm-dd format.
Documented here: https://perldoc.perl.org/Time/Piece.html
answered Jan 7 at 17:58
glenn jackmanglenn jackman
51k571110
51k571110
add a comment |
add a comment |
It could be useful to know which ksh88,93 are you using.
According to ksh93(1) you can find something about date:
printf format [ arg ... ]
%(date-format)T
--practical example
: printf "%(%d.%m,%Y)Tn"
11.01,2019
: printf "%(%d.%m.%Y)Tn" "2 days ago"
09.01.2019
Let's hope it helps you.
add a comment |
It could be useful to know which ksh88,93 are you using.
According to ksh93(1) you can find something about date:
printf format [ arg ... ]
%(date-format)T
--practical example
: printf "%(%d.%m,%Y)Tn"
11.01,2019
: printf "%(%d.%m.%Y)Tn" "2 days ago"
09.01.2019
Let's hope it helps you.
add a comment |
It could be useful to know which ksh88,93 are you using.
According to ksh93(1) you can find something about date:
printf format [ arg ... ]
%(date-format)T
--practical example
: printf "%(%d.%m,%Y)Tn"
11.01,2019
: printf "%(%d.%m.%Y)Tn" "2 days ago"
09.01.2019
Let's hope it helps you.
It could be useful to know which ksh88,93 are you using.
According to ksh93(1) you can find something about date:
printf format [ arg ... ]
%(date-format)T
--practical example
: printf "%(%d.%m,%Y)Tn"
11.01,2019
: printf "%(%d.%m.%Y)Tn" "2 days ago"
09.01.2019
Let's hope it helps you.
answered Jan 11 at 20:19
tntxtntx
642
642
add a comment |
add a comment |
Thanks for contributing an answer to Unix & Linux Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f492998%2fvariable-in-date-command%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
What language is
rel_date
a variable in?– Jesse_b
Jan 7 at 13:56
Will your input date always be in the same format? Also will it pad day numbers less than 10? ie
2019JAN07
or2019JAN7
?– Jesse_b
Jan 7 at 14:03
What OS and/or Distribution is this running on? (Do you have GNU
date
?)– roaima
Jan 7 at 14:08
3
"I have used various date command features" care to share which ones you've tried, and how? Or do we need to guess?
– roaima
Jan 7 at 14:08
2
"I am getting the error in stating -d option is not a valid argument." What command did you use, and what was the exact error message? (Don't approximate.)
– roaima
Jan 7 at 14:09