change format from %Y%m%d%H%M% to another format - linux
how can I change date format from "%Y%m%d%H%M" to "%Y-%m-%d" "%H:%M" in shell scripting?
there is a log file (logfile.txt) with date format as "%Y%m%d%H%M%S" in first column. what is needed is print first column like "%Y-%m-%d" "%H:%M".
e.g: from 201804041323 to 2018-04-04 13:23
Thanks in advance :)
perl -i.bak -pe 's/^(\d{4})(\d{2})(\d{2})(\d{2})(\d{2})/$1-$2-$3 $4:$5/' file
sed -r 's/^([0-9]{4})([0-9]{2})([0-9]{2})([0-9]{2})([0-9]{2})/\1-\2-\3 \4:\5/g' logfile.txt > newlogfile.txt
You can use below command for getting output in format of "2018-04-18 22:14"
date +%Y-%m-%d" "%H:%M
Related
Code for linux date YYY-M-D
In the linux date command, I use this code: TODAY=$(date +"%Y-%m-%d") sample output YYYY-MM-DD, such as 2016-09-05. If I want output date YYYY-M-D, such as 2016-9-5, what should I do with this code?
Do it: TODAY=$(date +"%Y-%-m-%-d") Explaining, the - will remove the leading zeros... Example: September month... %m returns 09 %-m returns 9
I am not user you can do that date, but you fix it afterwards:
TODAY=$(date +"%Y-%m-%d")
TODAY="${TODAY//-0/-}"
Replace Text with Formatted Text on Linux
I'm looking for a way to replace the text between positions 10 and 17 in a file with a Linux command or script. For example, I'd like to replace the date text 20140101 with 01/01/2014. I'm hoping this is something I can in a single command from the command line with maybe sed or awk?
Using sed, you can capture the first 9 chars in a capture group that would be placed as is. The remaining would be broken in 3 capture groups and re-arranged as you desire.
Something like:
sed -r 's#(.{9})(.{4})(.{2})(.{2})#\1\3/\4/\2#' file
If you are on a system that does not have GNU sed escape ( ) { } with \.
If all your strings are dates then the best is to use date command: $ date -d 20140101 +%m/%d/%Y 01/01/2014 $ date -d 20140923 +"%m-%d-%Y %a %b" 09-23-2014 Tue Sep It is especially great tool to translate number of seconds from 1970 (unix epoch) used in many log files: $ date --date='#1411199063' +"%m-%d-%Y %H:%M:%S" 09-20-2014 07:44:23
Command Line Date Formatting Assistance
I am running a report for my boss, and there were dates showing as DD/MM/YYYY (example: 18/06/2013). What can I use in my command to change this to MM/DD/YYYY? I'm thinking SED but I am not sure. Also, the date is the first element in the log I'm looking at.
$ echo 18/06/2013 | sed -r 's|(..)/(..)/|\2/\1/|' 06/18/2013
Change date format in first column using awk/sed
I have a shell script which is automatically ran each morning which appends that days results to a text file. The file should have todays date on the first column followed by results separated by commas. I use the command date +%x to get the day in the required format (dd/mm/yy). However on one computer date +%x returns mm/dd/yyyy ( any idea why this is the case?). I then sort the data in the file in date order. Here is a snippet of such a text file 29/11/12,9654.80,194.32,2.01,7.19,-7.89,7.65,7.57,3.98,9625.27,160.10,1.66,4.90,-4.79,6.83,4.84,3.54 03/12/12,5184.22,104.63,2.02,6.88,-6.49,7.87,6.67,4.10,5169.52,93.81,1.81,5.29,-5.45,7.87,5.37,4.10 04/12/12,5183.65,103.18,1.99,6.49,-6.80,8.40,6.66,4.38,5166.04,95.44,1.85,6.04,-6.49,8.40,6.28,4.38 11/07/2012,5183.65,102.15,1.97,6.78,-6.36,8.92,6.56,4.67,5169.48,96.67,1.87,5.56,-6.10,8.92,5.85,4.67 07/11/2012,5179.39,115.57,2.23,7.64,-6.61,8.83,7.09,4.62,5150.17,103.52,2.01,7.01,-6.08,8.16,6.51,4.26 11/26/2012,5182.66,103.30,1.99,7.07,-5.76,7.38,6.37,3.83,5162.81,95.47,1.85,6.34,-5.40,6.65,5.84,3.44 11/30/2012,5180.82,95.19,1.84,6.51,-5.40,7.91,5.92,4.12,5163.98,91.82,1.78,5.58,-5.07,7.05,5.31,3.65 Is it possible to change the date format for the latter four lines to the correct date format using awk or sed? I only wish to change the date format for those in the form mm/dd/yyyy to dd/mm/yy.
It looks like you're using two different flavors (versions) of date. To check which versions you've got, I think GNU date accepts the --version flag whereas other versions, like BSD/OSX will not accept this flag.
Since you may be using completely different systems, it's probably safest to avoid date completely and use perl to print the current date:
perl -MPOSIX -e 'print POSIX::strftime("%d/%m/%y", localtime) . "\n"'
If you are sure you have GNU awk on both machines, you could use it like this:
awk 'BEGIN { print strftime("%d/%m/%y") }'
To fix the file you've got, here's my take using GNU awk:
awk '{ print gensub(/^(..\/)(..\/)..(..,)/, "\\2\\1\\3", "g"); next }1' file
Or using sed:
sed 's/^\(..\/\)\(..\/\)..\(..,\)/\2\1\3/' file
Results:
29/11/12,9654.80,194.32,2.01,7.19,-7.89,7.65,7.57,3.98,9625.27,160.10,1.66,4.90,-4.79,6.83,4.84,3.54
03/12/12,5184.22,104.63,2.02,6.88,-6.49,7.87,6.67,4.10,5169.52,93.81,1.81,5.29,-5.45,7.87,5.37,4.10
04/12/12,5183.65,103.18,1.99,6.49,-6.80,8.40,6.66,4.38,5166.04,95.44,1.85,6.04,-6.49,8.40,6.28,4.38
07/11/12,5183.65,102.15,1.97,6.78,-6.36,8.92,6.56,4.67,5169.48,96.67,1.87,5.56,-6.10,8.92,5.85,4.67
11/07/12,5179.39,115.57,2.23,7.64,-6.61,8.83,7.09,4.62,5150.17,103.52,2.01,7.01,-6.08,8.16,6.51,4.26
26/11/12,5182.66,103.30,1.99,7.07,-5.76,7.38,6.37,3.83,5162.81,95.47,1.85,6.34,-5.40,6.65,5.84,3.44
30/11/12,5180.82,95.19,1.84,6.51,-5.40,7.91,5.92,4.12,5163.98,91.82,1.78,5.58,-5.07,7.05,5.31,3.65
This should work: sed -re 's/^([0-9][0-9])\/([0-9][0-9])\/[0-9][0-9]([0-9][0-9])(.*)$/\2\/\1\/\3\4/' It can be made smaller but I made it so it would be more obvious what it does (4 groups, just switching month/day and removing first two chars of the year). Tip: If you don't want to cat the file you could to the changes in place with sed -i. But be careful if you put a faulty expression in you might end up breaking your source file. NOTE: This assumes that IF the year is specified with 4 digits, the month/day is reversed.
This below command will do it.
Note:No matter how many number of lines are present in the file.this will just change the last 4 lines.
tail -r your_file| awk -F, 'NR<5{split($1,a,"/");$1=a[2]"/"a[1]"/"a[3];print}1'|tail -r
Well i could figure out some way without using pipes and using a single awk statement and this solution does need a tail command:
awk -F, 'BEGIN{cmd="wc -l your_file";while (cmd|getline tmp);split(tmp,x)}x[1]-NR<=4{split($1,a,"/");$1=a[2]"/"a[1]"/"a[3];print}1' your_file
Another solution:
awk -F/ 'NR<4;NR>3{a=$1;$1=$2;$2=a; print $1"/"$2"/" substr($3,3,2) substr($3,5)}' file
Using awk:
$ awk -F/ 'NR>3{x=$1;$1=$2;$2=x}1' OFS="/" file
By using the / as the delimiter, all you need to do is swap the 1st and 2nd fields which is done here using a temporary variable.
give the input format when converting a date
I read the date from a file to a variable. The date has the format ddmmyyyy. It has to be converted to yyyy-mm-dd
I already searched this forum and got this far :
date -d '$DATE' +%F
The problem is the input format is not recognised. Is there any way I can specify the input date format?
On an other forum I found : date -d "${OLD_DATE}" -D "%d%m%Y" +%F
where -D should specify the input format but this doesn't work. But -D is unknown.
thanks for the help and sorry for my English.
You could to it like this:
echo "DDMMYYYY" | awk 'BEGIN {OFS="-"} {print substr($1,5,4), substr($1,3,2), substr($1,1,2)}'
Output:
YYYY-MM-DD
Yes, date understands a lot of formats for -d, but when it's just 8 digits in a row, it interprets it as YYYYmmdd. I'm not sure if you can force it to read it differently, but in this case you can use a simple editor such as awk or sed instead:
$ OLD_DATE='08032011'
$ echo $OLD_DATE | sed -r 's/(.{2})(.{2})(.{4})/\3-\2-\1/'
2011-03-08
This will work on GNU sed. Note that it doesn't check the input (for brevity).