How can we get weekday based on given date in unix

Very simple. Just use the date command itself with correct options.

$ date -j -f '%m-%d-%Y' "09-01-2017" +'%A'Friday

bash shell datetime unix unix-timestamp

If you have your date like this:

d="09-01-2017"

you need to reformat it to "YYYY-MM-DD"

date -d $(echo $d|awk -F- '{print $3 "-" $1 "-" $2}') +%A # DOW

bash shell datetime unix unix-timestamp

DayOfWeek=$(date +%A)

This would yield the day of week monday-sundayIf your input date is strictly in the format MM-DD-YYYY, use the following

IFS='-' read -ra ADDR <<< "09-01-2017"formattedDate=${ADDR[2]}-${ADDR[0]}-${ADDR[1]}date -d  $formattedDate +%A

The first line tokenizes the components of the date and the second rearranges them

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