I have a string '2015_11_01_00_00' I want this convert into epoch time.
I have tried the below code, but could not found it worth.
#!/bin/bash
echo "mohit"
date_var="2015_11_01_00_00"
echo $date_var
arr=$(echo $date_var | tr "_" "\n")
IFS='_ ' read -r -a arr <<< "$date_var"
#echo $arr
echo ${arr[0]}
echo ${arr[1]}
echo ${arr[2]}
echo ${arr[3]}
echo ${arr[4]}
updated_date="${arr[0]}"-"${arr[1]}"-"${arr[2]}"-"${arr[3]}"-"${arr[4]}"
echo $updated_date
TMRW_DATE=`date --date="$updated_date" +"%s"`
echo "$TMRW_DATE"
This one should do the trick:
#!/bin/bash
echo "mohit"
date_var="2015_11_01_00_00"
updated_date=`echo ${date_var} | sed -e 's,_,/,;s,_,/,;s,_, ,;s/_/:/'`
echo ${updated_date}
TMRW_DATE=`date -d "$updated_date" +"%s"`
echo "$TMRW_DATE"
Here is one way to do it, using GNU date syntax:
#!/bin/bash
echo "mohit"
date_var="2015_11_01_00_00"
TMRW_DATE=$(
set -- $(echo $date_var | tr "_" " ")
date --date="$1-$2-$3 $4:$5" +"%s"
)
echo "$TMRW_DATE"
and this should work with OS X date implementation :
#!/bin/bash
echo "mohit"
date_var="2015_11_01_00_00"
TMRW_DATE=$(date -j -f "%Y_%m_%d_%H_%M" "$date_var" +%s)
echo "$TMRW_DATE"
i was trying to do the same for postgres sql by following which produce error
SELECT *, DATEDIFF(ss, '1970-01-01T00:00:00.000', punch_time) as EpochSeconds
FROM attendance
but following code did work for me
SELECT EXTRACT(EPOCH FROM punch_time) as EpochTime
FROM attendance;
Related
I have date and time format like this(yearmonthday):
20141105 11:30:00
I need assignment year, month, day, hour and minute values to variable.
I can do it year, day and hour like this:
year=$(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.
So, everything works fine in the code, except for one tiny little thing.
This part:
if [ "$LIMITHOURS" -gt "0" -a "$LIMITHOURS" -lt "24" ]; then
x=$(($LIMITHOURS*60*60))
fi
SDATE=$( echo "01/jan/2003:11:00:06 +0100"| sed 's/[/]/ /g' |sed 's/:/ /')
EDATE=$(date --date "$SDATE - $x seconds" +"%d%m%Y%H%M%S")
#echo "$SDATE"
#echo "$EDATE"
while read LINE; do
CDATE=$( awk '{print $4}'| sed 's/[[]//' | sed 's/[/]//g' |sed 's/://g' )
DATE=$(date --date "$CDATE" +"%d%m%Y%H%M%S")
#echo "$CDATE"
done < "$FILENAME"
When I try to run the script, I get the error message "date: Argument list too long
" and I know that the problem is in the while loop, with:
DATE=$(date --date "$CDATE" +"%d%m%Y%H%M%S")
Anyone who know any solution for this? I want the date format in ddmmYYYYHHMMSS, eg. 23102002120022
You can find rest of the script here: http://pastebin.com/PMk2QDre
This code:
while read LINE; do
CDATE=$( awk '{print $4}'| sed 's/[[]//' | sed 's/[/]//g' |sed 's/://g' )
DATE=$(date --date "$CDATE" +"%d%m%Y%H%M%S")
#echo "$CDATE"
done < "$FILENAME"
will read one line from $FILENAME into the variable LINE, but then the first call to awk is reading the rest of the lines. The resulting CDATE value is probably too large to fit in a single command line, never mind it containing too many dates. You probably wanted
echo "$LINE" | awk '{print $4}' | ...
A simpler way to strip the undesirable characters from LINE, however, is
CDATE=${LINE//[\/[:]}
past_date='2013-11-14'
initial_time=$(grep -o -m1 "$past_date [0-9][0-9]:[0-9][0-9]:[0-9][0-9]" logfile.txt)
/* Here I am trying to find the first occurence of date which is greater than or eqaul to '2013-11-14', Above code I have tried ,It is giving only that particular line of file, If that date is not found It has to give next date which is greater than 2013-11-14 date */
Using awk
past_date='20131114'
awk '{d=$1;gsub(/-/,"",d);if (d>=p) {print;exit}}' p=$past_date logfile
2013-11-15 15:45:40 Starting agent install process
If you use bash, then you might want to try something like:
past_date='2013-11-14'
initial_time=$(grep -oP '\d{4}-\d\d-\d\d \d\d:\d\d:\d\d' < logfile.txt | \
while read LINE ; do if [ "$LINE" '>' "$past_date" ]; then echo $LINE; break; fi ; done)
while read line
do
initial_time=`echo $line | sed -e 's/\([0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9] [0-9][0-9]:[0-9][0-9]:[0-9][0-9]\).*/\1/'`
file_content_date=`date -d "$initial_time" +%Y%m%d`
comparison_past_date=`date -d "$past_date" +%Y%m%d`
if [ $comparison_past_date -le $file_content_date ]; then
comparison_start_date=`date -d "$file_content_date" +%Y%m%d`
break
fi
done < logfile.txt
fi
I have a file of filenames, and I need to be able to get the size of these files using bash.
I have the following script which does that, but It prints the filename and the size on different lines, i'd prefer it to do it all on one line if possible.
#!/bin/sh
filename="$1"
while read -r line
do
name=$line
vars=(`echo $name | tr '.' ' '`)
echo $name
stat -x $name | grep Size: | awk '{ print $2 }'
done < "$filename"
I'd love to have it of the form:
filename: $size
How can I do this?
(I am using OSX hence the slightly odd version of stat.)
Pass -n to the echo to prevent a trailing newline from being added. So change
echo $name
to
echo -n $name
and to add the : separator between the file name and file size
echo -n ${name}": "
This should do the trick:
while read f
do
echo "${f} : $(stat -L -c %s ${f})"
done < "${filename}"
echo $name: $(stat -x $name | sed -n '/^Size:/s///p')
I have in a bash script:
for i in `seq 1 10`
do
read AA BB CC <<< $(cat file1 | grep DATA)
echo ${i}
echo ${CC}
SORT=${CC}${i}
echo ${SORT}
done
so "i" is a integer, and CC is a string like "TODAY"
I would like to get then in SORT, "TODAY1", etc
But I get "1ODAY", "2ODAY" and so
Where is the error?
Thanks
You should try
SORT="${CC}${i}"
Make sure your file does not contain "\r" that would end just in the end of $CC.
This could well explain why you get "1ODAY".
Try including
|tr '\r' ''
after the cat command
try
for i in {1..10}
do
while read -r line
do
case "$line" in
*DATA* )
set -- $line
CC=$3
SORT=${CC}${i}
echo ${SORT}
esac
done <"file1"
done
Otherwise, show an example of file1 and your desired output
ghostdog is right: with the -r option, read avoids succumbing to potential horrors, like CRLFs. Using arrays makes the -r option more pleasant:
for i in `seq 1 10`
do
read -ra line <<< $(cat file1 | grep DATA)
CC="${line[3]}"
echo ${i}
echo ${CC}
SORT=${CC}${i}
echo ${SORT}
done