date -d 09/20/1981 +"%Y-%m-%d"

Change date from MM/DD/YYYY to YYYY-MM-DD (mysql like)


0
By: trunet
2009-02-09 18:30:44

These Might Interest You

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    1
    YYYY=2014; MM=02; for d in $(cal -h $MM $YYYY | grep "^ *[0-9]"); do DD=$(printf "%02d" $d); echo $YYYY$MM$DD; done
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  • * Replace USERNAME with the desired svn username * Replace the first YYYY-MM-DD with the date you want to get the log (this starts at the midnight event that starts this date) * Replace the second YYYY-MM-DD with the date after you want to get the log (this will end the log scan on midnight of the previous day) Example, if I want the log for December 10, 2010, I would put {2010-12-10}:{2010-12-11} Show Sample Output


    1
    svn log -r '{YYYY-MM-DD}:{YYYY-MM-DD}' | sed -n '1p; 2,/^-/d; /USERNAME/,/^-/p' | grep -E -v '^(r[0-9]|---|$)' | sed 's/^/* /g'
    antic · 2010-12-22 17:52:19 1
  • watch the seconds of your life tick away - replace YYYY-mm-dd HH:MM:ss w/ your birthtime.


    1
    while [ 0 ]; do expr 2365200000 \- `date +%s` \- `date --date "YYYY-mm-dd HH:MM:ss" +%s`; sleep 1; clear; done
    wwest4 · 2009-02-13 20:02:37 0
  • On CentOS at least, date returns a boolean for the common date string formats, including YYYY-MM-DD. In the sample output, you can see various invalid dates returning 0 whereas a simple regex check would return 1 for the invalid dates. -d, --date=STRING display time described by STRING, not `now' The version of date on OS X does not appear to have this same option. Show Sample Output


    0
    if date -d 2006-10-10 >> /dev/null 2>&1; then echo 1; else echo 0; fi
    SteveGoossens · 2013-01-10 10:35:15 0
  • Print out your age in days in binary. Today's my binary birthday, I'm 2^14 days old :-) . This command does bash arithmatic $(( )) on two dates: Today: $(date +%s) Date of birth: $(date +%s -d YYYY-MM-DD) The dates are expressed as the number of seconds since the Unix epoch (Jan 1970), so we devide the difference by 86400 (seconds per day). . Finally we pipe "obase=2; DAYS-OLD" into bc to convert to binary. (obase == output base) Show Sample Output


    2
    echo "obase=2;$((($(date +%s)-$(date +%s -d YYYY-MM-DD))/86400))" | bc
    flatcap · 2015-10-19 15:40:32 0
  • Quick and easy way of validating a date format of yyyy-mm-dd and returning a boolean, the regex can easily be upgraded to handle "in betweens" for mm dd or to validate other types of strings, ex. ip address. Boolean output could easily be piped into a condition for a more complete one-liner.


    -1
    echo 2006-10-10 | grep -c '^[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9]$'
    rez0r · 2009-05-11 22:18:43 3

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