Convert date formats in bash

Question

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

User · Answer

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 lt 11  i       do             2     months  i       amp  amp  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

User · Answer

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

User · Answer

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

User · Answer

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  lt input format gt   lt date gt   lt output format gt      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    gt   2017-12-10  convert date format   b  d  T  Y  Z   Dec 10 17 30 05 2017 GMT    Y- m- d

User · Answer

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  gt  gt    DATE  printf    AS YYYYMMDD format  gt  gt   s    date -d  DATE    Y m d        which then outputs the following   Get date  gt  gt  27 june 2011     AS YYYYMMDD format  gt  gt  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

User · Answer

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

User · Answer

date -d  25 JUN 2011    Y m d   outputs  20110625

[linux] Convert date formats in bash

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