I need to convert the date of the macbook via CLI because I'm running a script:
Date we have to convert 2019-09-17 15:32:27
Final format Jun 17 15:32:27
I tried it with the following script:
date_raw=`grep "^.*,.*,.*,.*,.*,.*,"$i"," ${SNIFFER_BUFFER_USAGE} | head -1 | awk -F ',' '{ print $2}'`
if [ $OSX -eq 1 ];
then
dataHoraMinuto=$(date -j -f "%Y-%m-%d %H:%M:%S" "$date_raw" +"%b %d %H:%M:%S")
else
dataHoraMinuto=`date +"%b %d %H:%M:%S" -d "$date_raw"`
fi
$ date -d"2019-09-17 15:32:27" +"%b %d %T"
Sep 17 15:32:27
Related
I have a date in this format: "27 JUN 2011" and I want to convert it to 20110627
Is it possible to do in bash?
#since this was yesterday
date -dyesterday +%Y%m%d
#more precise, and more recommended
date -d'27 JUN 2011' +%Y%m%d
#assuming this is similar to yesterdays `date` question from you
#http://stackoverflow.com/q/6497525/638649
date -d'last-monday' +%Y%m%d
#going on #seth's comment you could do this
DATE="27 jun 2011"; date -d"$DATE" +%Y%m%d
#or a method to read it from stdin
read -p " Get date >> " DATE; printf " AS YYYYMMDD format >> %s" `date
-d"$DATE" +%Y%m%d`
#which then outputs the following:
#Get date >> 27 june 2011
#AS YYYYMMDD format >> 20110627
#if you really want to use awk
echo "27 june 2011" | awk '{print "date -d\""$1FS$2FS$3"\" +%Y%m%d"}' | bash
#note | bash just redirects awk's output to the shell to be executed
#FS is field separator, in this case you can use $0 to print the line
#But this is useful if you have more than one date on a line
More on Dates
note this only works on GNU date
I have read that:
Solaris version of date, which is unable
to support -d can be resolve with
replacing sunfreeware.com version of
date
On OSX, I'm using -f to specify the input format, -j to not attempt to set any date, and an output format specifier. For example:
$ date -j -f "%m/%d/%y %H:%M:%S %p" "8/22/15 8:15:00 am" +"%m%d%y"
082215
Your example:
$ date -j -f "%d %b %Y" "27 JUN 2011" +%Y%m%d
20110627
date -d "25 JUN 2011" +%Y%m%d
outputs
20110625
If you would like a bash function that works both on Mac OS X and Linux:
#
# Convert one date format to another
#
# Usage: convert_date_format <input_format> <date> <output_format>
#
# Example: convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'
convert_date_format() {
local INPUT_FORMAT="$1"
local INPUT_DATE="$2"
local OUTPUT_FORMAT="$3"
local UNAME=$(uname)
if [[ "$UNAME" == "Darwin" ]]; then
# Mac OS X
date -j -f "$INPUT_FORMAT" "$INPUT_DATE" +"$OUTPUT_FORMAT"
elif [[ "$UNAME" == "Linux" ]]; then
# Linux
date -d "$INPUT_DATE" +"$OUTPUT_FORMAT"
else
# Unsupported system
echo "Unsupported system"
fi
}
# Example: 'Dec 10 17:30:05 2017 GMT' => '2017-12-10'
convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'
Just with bash:
convert_date () {
local months=( JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC )
local i
for (( i=0; i<11; i++ )); do
[[ $2 = ${months[$i]} ]] && break
done
printf "%4d%02d%02d\n" $3 $(( i+1 )) $1
}
And invoke it like this
d=$( convert_date 27 JUN 2011 )
Or if the "old" date string is stored in a variable
d_old="27 JUN 2011"
d=$( convert_date $d_old ) # not quoted
It's enough to do:
data=`date`
datatime=`date -d "${data}" '+%Y%m%d'`
echo $datatime
20190206
If you want to add also the time you can use in that way
data=`date`
datatime=`date -d "${data}" '+%Y%m%d %T'`
echo $data
Wed Feb 6 03:57:15 EST 2019
echo $datatime
20190206 03:57:15
Maybe something changed since 2011 but this worked for me:
$ date +"%Y%m%d"
20150330
No need for the -d to get the same appearing result.
For example:
echo $(date) - $(date -r sample.txt)
Output:
90 days(for example)
Use %s seconds since 1970-01-01 00:00:00 UTC as in
echo $(expr $(date +%s) - $(date -r sample.txt +%s)) #!/bin/sh
echo $(($(date +%s) - $(date -r sample.txt +%s))) #/bin/bash
One more way
$ ls -l peter.txt
-rwxrw-r--+ 1 pppp qqqq 149 Dec 15 18:39 peter.txt
$ echo "(" $(date +%s) - $(date -r peter.txt +%s) ")/" 86400 | perl -nle ' print eval, " days" '
29.254537037037 days
$
How to change this time output ?
date --date="#$(echo $(TZ=UTC date +%s) - $(date +%s)'%(5*60)-(5*60)' | bc)"
Output: za apr 12 00:25:00 CEST 2014
Should output in this layout: %Y%m%d%H%M
How to implement this in the string ?
Thanks!!
i think this should do what you want:
d="#$(echo $(TZ=UTC date +%s) - $(date +%s)'%(5*60)-(5*60)' | bc)"&&echo `date --date="$d"` `date --date="$d" +%Y%m%d%H%M`
d="#$(echo $(TZ=UTC date +%s) - $(date +%s)'%(5*60)-(5*60)' | bc)"&&echo `date --date="$d" --utc` `date --date="$d" +%Y%m%d%H%M --utc`
Second one is in UTC.
I have tried adding -d "yesterday" but I haven't had any luck getting it to work. Here is what I have for the whole script:
#! /bin/bash
saveDir="TJ"
dd=$(date +"%m-%d-%Y")
for file in *.csv ; do
saveName="${saveDir}/TJ ${dd}.csv"
cut -d',' -f2,14 "$file" > "$saveName"
done
how do I get dd to output yesterdays date instead of the current date?
EDIT: This is what I have now
#! /bin/bash
saveDir="TJ"
dd=$(date --date='yesterday' +'%m-%d-%Y')
for file in *.csv ; do
saveName="${saveDir}/TJ ${dd}.csv"
cut -d',' -f2,14 "$file" > "$saveName"
done
the above is saving the file as TJ .csv but I'm not sure what was done incorrectly
I think you want to use -
$ cat test.sh
#!/bin/bash
dd=$(date --date='yesterday' +'%m-%d-%Y')
echo $dd
$ ./test.sh
12-31-2013
or you could use
$ date -d '1 day ago' +'%m-%d-%Y'
12/31/2013
And for tomorrow -
$ date -d '1 day' +'%m-%d-%Y'
01/02/2014
or
$ date --date='tomorrow'
Thu Jan 2 21:25:00 EST 2014
Get today's date in seconds since epoch. Subtract 86400 to get to yesterday. Then convert yesterday to the string format you want.
today=`date +"%s"`
yesterday=`expr $today - 86400`
dd=`date --date="#${yesterday}" +"%m-%d-%Y"`
Try this
yday=$(date --date yesterday "+%d-%m-%Y")
echo $yday
And If you works in Linux
yday=$(date -d "-1 days" +"%d-%m-%Y")
echo $yday
I tried date -d "yesterday" +%m-%d-%Y on my Linux, it worked fine.
If you are on unix platform, you cannot use -d,
you can get yesterday's date using perl, this is how I do using perl
dd=$(perl -e '($a,$b,$c,$day,$mon,$year,$d,$e,$f) = localtime(time-86400);printf "%02d-%02d-%4d",$day, $mon+1, $year+1900')
echo $dd
01-01-2014
NewDate=`date +"%A %d %B %Y" --date="-1 day"`
echo $NewDate
this will give your yesterday's date (-1)
This will give you tomorrow's date (+1)
even you can check for any values like (+/-) days
I am running a shell script which accepts date in "YYYY-MM-DD" format. from this date input, how can i get year, month and day separately?
Thanks for replies in advance.
except for processing the string as text(with grep/sed/awk/cut/...), you could do with with date command:
kent$ date -d '2013-09-06' +%Y
2013
kent$ date -d '2013-09-06' +%m
09
kent$ date -d '2013-09-06' +%d
06
You could do this to store them on variables with one command:
read YEAR MONTH DAY < <(date -d '2013-09-06' '+%Y %m %d')
printf "%s\n" "$YEAR" "$MONTH" "$DAY"
Try this :
dt="2011-2-3"
arr=( $( date --date=$dt "+%Y %m %d") )
echo "Year > $arr"
echo "Month > ${arr[1]}"
echo "Month > ${arr[2]}"
I didn't want to use date command, so I got the following code. It uses awk, which I feel is more appropriate for this task. I assumed you want to store results in variables.
inputDate=2017-07-06
year=`echo $inputDate | awk -F\- '{print $1}'`
month=`echo $inputDate | awk -F\- '{print $2}'`
day=`echo $inputDate | awk -F\- '{print $3}'`