I need help in completing the below script in shell.
Requested format is to print previous, current and next business days (should exclude weekends and holidays).
text name : holiday.txt
#!/usr/bin/sh
cur_date=`date +"%A %m/%d/%Y"`
cur_day=`date +"%A"`
yesterday=`TZ=AEDT+13 date +"%A %m/%d/%Y"`;
twodd_back=`TZ=AEDT+37 date +"%A %m/%d/%Y"`;
tomorrow=`TZ=AEDT-35 date +"%A %m/%d/%Y"`;
twodd_later=`TZ=AEDT-59 date +"%A %m/%d/%Y"`;
cdate=`date +"%m/%d/%Y"`
####################
echo "Previous Business Day is $yesterday"
if echo $yesterday|awk '{print $1}'=Saturday
then echo "Previous Business Day is $twodd_back"
fi
echo "Current Business Day is $ cur_date"
echo " Next Business Day is $tomorrow"
The script in its current status does not execute successfully. It fails with below error.
awk: syntax error at source line 1
context is
{print >>> $1}= <<< Saturday
awk: bailing out at source line 1
To fix it, change your if statement.
if [[ `echo $yesterday | awk '{print $1}'` = Saturday ]];
Related
I am looking to write a shell script that finds and replaces the parameter values every time a workflow is run. I am a beginner and trying to learn more about this.
For example
-$$mp_Custom_Filter_D_Prem=AND BLSB.BLSB_CREATE_DTM <= '2019-02-28'
--$$mp_Custom_Filter_D_LEP=AND convert(date,(substring(SBSR.SBSR_SOURCE,9,2)+substring(SBSR.SBSR_SOURCE,5,2)+substring(SBSR.SBSR_SOURCE,7,2))) <= '2019-02-28'
I have these 2 parameters in the file and I want the script to change the dates to advance a month and set its last day.
So far the date 2019-02-28 I would like the script to find the dates and replace it to 2019-03-31.
I have tried the below script and its not the same result
PRM_FIL_DIR=$1 PRM_FIL_NME=$2 LOG_FIL_DIR=$3 LOG_FIL_NME=$4 echo Begining of the log File > $LOG_FIL_DIR/$LOG_FIL_NME From=`grep '$$mp_Custom_Filter_D_Prem=AND BLSB.BLSB_CREATE_DTM <' $PRM_FIL_DIR/$PRM_FIL_NME | awk -F= '{print $3}'|uniq| sed "s/'//g" | awk '$1=$1' ` echo $From is Date value found in $PRM_FIL_DIR/$PRM_FIL_NME parameter File >> $LOG_FIL_DIR/$LOG_FIL_NME To=`date -d "$From 2 month -1 day" +%Y-%m-%d` echo $To is the value to be replaced in $PRM_FIL_DIR/$PRM_FIL_NME parameter File >> $LOG_FIL_DIR/$LOG_FIL_NME From_FNL="'$From'" To_FNL="'$To'" sed -i "s/$From_FNL/$To_FNL/g" "$PRM_FIL_DIR/$PRM_FIL_NME" echo Date values replaced >> $LOG_FIL_DIR/$LOG_FIL_NME echo End of the log File >> $LOG_FIL_DIR/$LOG_FIL_NME PRM_FIL_DIR=$1 PRM_FIL_NME=$2 LOG_FIL_DIR=$3 LOG_FIL_NME=$4 echo Begining of the log File > $LOG_FIL_DIR/$LOG_FIL_NME From=`grep '$$mp_Custom_Filter_D_Prem=AND BLSB.BLSB_CREATE_DTM <' $PRM_FIL_DIR/$PRM_FIL_NME | awk -F= '{print $3}'|uniq| sed "s/'//g" | awk '$1=$1' ` echo $From is Date value found in $PRM_FIL_DIR/$PRM_FIL_NME parameter File >> $LOG_FIL_DIR/$LOG_FIL_NME To=`date -d "$From 2 month -1 day" +%Y-%m-%d` echo $To is the value to be replaced in $PRM_FIL_DIR/$PRM_FIL_NME parameter File >> $LOG_FIL_DIR/$LOG_FIL_NME From_FNL="'$From'" To_FNL="'$To'" sed -i "s/$From_FNL/$To_FNL/g" "$PRM_FIL_DIR/$PRM_FIL_NME" echo Date values replaced >> $LOG_FIL_DIR/$LOG_FIL_NME echo End of the log File >> $LOG_FIL_DIR/$LOG_FIL_NME
I want the output to be the last day of the month i.e 2019-03-31 and actual output is 2019-03-29
Try this flow,
lastmonth=$( date -d "-$(date +%d) days" +%Y%m%d)
currentmonth=$( date -d "-$(date +%d) days month" +%Y%m%d)
sed "s/$lastmonth/$currentmonth/g" filename.txt
I have date and time format like this(yearmonthday):
20141105 11:30:00
I need assignment year, month, day, hour and minute values to variable.
I can do it year, day and hour like this:
year=$(awk '{print $1}' log.log | sed 's/^\(....\).*/\1/')
day=$(awk '{print $1}' log.log | sed 's/^.*\(..\).*/\1/')
hour=$(awk '{print $2}' log.log | sed 's/^\(..\).*/\1/')
How can I do this for month and minute?
--
And I need that every line of my log file:
20141105 11:30:00 /bla/text.1
20141105 11:35:00 /bla/text.2
20141105 11:40:00 /bla/text.3
....
I'm trying read line by line this log file and do this:
mkdir -p "/bla/backup/$year/$month/$day/$hour/$minute"
mv $file "/bla/backup/$year/$month/$day/$hour/$minute"
Here is my not working code:
#!/bin/bash
LOG=/var/log/LOG
while read line
do
year=${line:0:4}
month=${line:4:2}
day=${line:6:2}
hour=${line:9:2}
minute=${line:12:2}
file=$(awk '{print $3}')
if [ -f "$file" ]; then
printf -v path "%s/%s/%s/%s/%s" $year $month $day $hour $minute
mkdir -p "/bla/backup/$path"
mv $file "/bla/backup/$path"
fi
done < $LOG
You don't need to call out to awk to date at all, use bash's substring operations
d="20141105 11:30:00"
yr=${d:0:4}
mo=${d:4:2}
dy=${d:6:2}
hr=${d:9:2}
mi=${d:12:2}
printf -v dir "/bla/%s/%s/%s/%s/%s/\n" $yr $mo $dy $hr $mi
echo "$dir"
/bla/2014/11/05/11/30/
Or directly, without all the variables.
printf -v dir "/bla/%s/%s/%s/%s/%s/\n" ${d:0:4} ${d:4:2} ${d:6:2} ${d:9:2} ${d:12:2}
Given your log file:
while read -r date time file; do
d="$date $time"
printf -v dir "/bla/%s/%s/%s/%s/%s/\n" ${d:0:4} ${d:4:2} ${d:6:2} ${d:9:2} ${d:12:2}
mkdir -p "$dir"
mv "$file" "$dir"
done < filename
or, making a big assumption that there are no whitespace or globbing characters in your filenames:
sed -r 's#(....)(..)(..) (..):(..):.. (.*)#mv \6 /blah/\1/\2/\3/\4/\5#' | sh
date command also do this work
#!/bin/bash
year=$(date +'%Y' -d'20141105 11:30:00')
day=$(date +'%d' -d'20141105 11:30:00')
month=$(date +'%m' -d'20141105 11:30:00')
minutes=$(date +'%M' -d'20141105 11:30:00')
echo "$year---$day---$month---$minutes"
You can use only one awk
month=$(awk '{print substr($1,5,2)}' log.log)
year=$(awk '{print substr($1,0,4)}' log.log)
minute=$(awk '{print substr($2,4,2)}' log.log)
etc
I guess you are processing the log file, which each line starts with the date string. You may have already written a loop to handle each line, in your loop, you could do:
d="$(awk '{print $1,$2}' <<<"$line")"
year=$(date -d"$d" +%Y)
month=$(date -d"$d" +%m)
day=$(date -d"$d" +%d)
min=$(date -d"$d" +%M)
Don't repeat yourself.
d='20141105 11:30:00'
IFS=' ' read -r year month day min < <(date -d"$d" '+%Y %d %m %M')
echo "year: $year"
echo "month: $month"
echo "day: $day"
echo "min: $min"
The trick is to ask date to output the fields you want, separated by a character (here a space), to put this character in IFS and ask read to do the splitting for you. Like so, you're only executing date once and only spawn one subshell.
If the date comes from the first line of the file log.log, here's how you can assign it to the variable d:
IFS= read -r d < log.log
eval "$(
echo '20141105 11:30:00' \
| sed 'G;s/\(....\)\(..\)\(..\) \(..\):\(..\):\(..\) *\(.\)/Year=\1\7Month=\2\7Day=\3\7Hour=\4\7Min=\5\7Sec=\6/'
)"
pass via a assignation string to evaluate. You could easily adapt to also check the content by replacing dot per more specific pattern like [0-5][0-9] for min and sec, ...
posix version so --posix on GNU sed
I wrote a function that I usually cut and paste into my script files
function getdate()
{
local a
a=(`date "+%Y %m %d %H %M %S" | sed -e 's/ / /'`)
year=${a[0]}
month=${a[1]}
day=${a[2]}
hour=${a[3]}
minute=${a[4]}
sec=${a[5]}
}
in the script file, on a line of it's own
getdate
echo "year=$year,month=$month,day=$day,hour=$hour,minute=$minute,second=$sec"
Of course, you can modify what I provided or use answer [6] above.
The function takes no arguments.
I am sure there is a simple answer to this :
Why does the following not give me the date and time in a script? What should I use rather than awk? Awk works fine if I need to query files but I guess it is the wrong thing to use here..
Code :
$ySTD=$(date --date yesterday "+%Y-%m-%d") | awk '{print substr($1,1,10)}'
echo ${ySTD}
echo 'STD calculated from DATE - last 24 hrs data will be fetched'
${ySTM}=$(date --date yesterday "+%T")| awk '{print substr($1,1,5)}'
echo ${ySTM}
In the shell I get this when the script runs:
+ awk '{print substr($1,1,10)}'
++ date --date yesterday +%Y-%m-%d
+ =2014-07-03
logdat3: line 41: =2014-07-03: command not found
+ echo
+ echo 'STD calculated from DATE - last 24 hrs data will be fetched'
STD calculated from DATE - last 24 hrs data will be fetched
+ awk '{print substr($1,1,5)}'
++ date --date yesterday +%T
+ =11:33:34
logdat3: line 45: =11:33:34: command not found
+ echo
Many thanks in advance
You're rigth, no awk needed:
$ read ySTD ySTM < <(date --date yesterday "+%Y-%m-%d %T")
$ echo $ySTD
2014-07-03
$ echo $ySTM
12:16:37
Anyway , your original code fixed:
ySTD=$(date --date yesterday "+%Y-%m-%d")
echo ${ySTD}
echo 'STD calculated from DATE - last 24 hrs data will be fetched'
ySTM=$(date --date yesterday "+%T")
echo ${ySTM}
You don't need to get a substring.
I am running a shell script which accepts date in "YYYY-MM-DD" format. from this date input, how can i get year, month and day separately?
Thanks for replies in advance.
except for processing the string as text(with grep/sed/awk/cut/...), you could do with with date command:
kent$ 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}'`
I tried using $(date) in my bash shell script, however, I want the date in YYYY-MM-DD format.
How do I get this?
In bash (>=4.2) it is preferable to use printf's built-in date formatter (part of bash) rather than the external date (usually GNU date).
As such:
# put current date as yyyy-mm-dd in $date
# -1 -> explicit current date, bash >=4.3 defaults to current time if not provided
# -2 -> start time for shell
printf -v date '%(%Y-%m-%d)T\n' -1
# put current date as yyyy-mm-dd HH:MM:SS in $date
printf -v date '%(%Y-%m-%d %H:%M:%S)T\n' -1
# to print directly remove -v flag, as such:
printf '%(%Y-%m-%d)T\n' -1
# -> current date printed to terminal
In bash (<4.2):
# put current date as yyyy-mm-dd in $date
date=$(date '+%Y-%m-%d')
# put current date as yyyy-mm-dd HH:MM:SS in $date
date=$(date '+%Y-%m-%d %H:%M:%S')
# print current date directly
echo $(date '+%Y-%m-%d')
Other available date formats can be viewed from the date man pages (for external non-bash specific command):
man date
Try: $(date +%F)
The %F option is an alias for %Y-%m-%d
You can do something like this:
$ date +'%Y-%m-%d'
$(date +%F)
output
2018-06-20
Or if you also want time:
$(date +%F_%H-%M-%S)
can be used to remove colons (:) in between
output
2018-06-20_09-55-58
You're looking for ISO 8601 standard date format, so if you have GNU date (or any date command more modern than 1988) just do: $(date -I)
date -d '1 hour ago' '+%Y-%m-%d'
The output would be 2015-06-14.
With recent Bash (version ≥ 4.2), you can use the builtin printf with the format modifier %(strftime_format)T:
$ printf '%(%Y-%m-%d)T\n' -1 # Get YYYY-MM-DD (-1 stands for "current time")
2017-11-10
$ printf '%(%F)T\n' -1 # Synonym of the above
2017-11-10
$ printf -v date '%(%F)T' -1 # Capture as var $date
printf is much faster than date since it's a Bash builtin while date is an external command.
As well, printf -v date ... is faster than date=$(printf ...) since it doesn't require forking a subshell.
I use the following formulation:
TODAY=`date -I`
echo $TODAY
Checkout the man page for date, there is a number of other useful options:
man date
if you want the year in a two number format such as 17 rather than 2017, do the following:
DATE=`date +%d-%m-%y`
I use $(date +"%Y-%m-%d") or $(date +"%Y-%m-%d %T") with time and hours.
I used below method. Thanks for all methods/answers
ubuntu#apj:/tmp$ datevar=$(date +'%Y-%m-%d : %H-%M')
ubuntu#apj:/tmp$ echo $datevar
2022-03-31 : 10-48
Whenever I have a task like this I end up falling back to
$ man strftime
to remind myself of all the possibilities for time formatting options.
Try to use this command :
date | cut -d " " -f2-4 | tr " " "-"
The output would be like: 21-Feb-2021
#!/bin/bash -e
x='2018-01-18 10:00:00'
a=$(date -d "$x")
b=$(date -d "$a 10 min" "+%Y-%m-%d %H:%M:%S")
c=$(date -d "$b 10 min" "+%Y-%m-%d %H:%M:%S")
#date -d "$a 30 min" "+%Y-%m-%d %H:%M:%S"
echo Entered Date is $x
echo Second Date is $b
echo Third Date is $c
Here x is sample date used & then example displays both formatting of data as well as getting dates 10 mins more then current date.
You can set date as environment variable and later u can use it
setenv DATE `date "+%Y-%m-%d"`
echo "----------- ${DATE} -------------"
or
DATE =`date "+%Y-%m-%d"`
echo "----------- ${DATE} -------------"
echo "`date "+%F"`"
Will print YYYY-MM-DD
Try this code for a simple human readable timestamp:
dt=$(date)
echo $dt
Output:
Tue May 3 08:48:47 IST 2022