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?

Answer 1

date -d "25 JUN 2011" +%Y%m%d

outputs

20110625

Answer 2

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.

Answer 3

#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 >> " DATE; printf "  AS YYYYMMDD format >> %s"  `date
-d"$DATE" +%Y%m%d`    

#which then outputs the following:
#Get date >> 27 june 2011   
#AS YYYYMMDD format >> 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

Answer 4

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<11; i++ )); do
        [[ $2 = ${months[$i]} ]] && 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

Answer 5

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 <input_format> <date> <output_format>
#
# 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' => '2017-12-10'
convert_date_format '%b %d %T %Y %Z' 'Dec 10 17:30:05 2017 GMT' '%Y-%m-%d'

Answer 6

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