Get date with same day in month - linux

Question

I want to get all dates with the same day of week.
inputDate="2021/08/25"
That means I should get all the same day of week as inputDate.
outputDates="2021/08/04,2021/08/11,2021/08/18,2021/08/25"
I only got this so far..
inputDate="2021/08/25"
dd=$(date -d "$inputDate" +"%Y/%m/%d")
So what I'm planning is to do "date -7" and loop 5 times forward and backward and collect it then check if value of month is still the same with inputDate if not then drop it
Do you have any way to do this?

Answer 1

Using only shell, the easyest way to get all weekdays from a month is by using cal command:
cal -n1 8 2021
outputs:
August 2021
Su Mo Tu We Th Fr Sa
1 2 3 4 5 6 7
8 9 10 11 12 13 14
15 16 17 18 19 20 21
22 23 24 25 26 27 28
29 30 31
Then you can filter using sed, awk or other tools to reach your goal.
Example:
year=2021
month=8
day=25
weekday_number="$(date -d$year-$month-$day +%w)"
cn=$(($weekday_number + 1))
cal -n1 $month $year |
sed -r 's/(..)\s/\1\t/g;s/ +//g' |
awk -v cn=$cn -F'\t' 'NR<3 || $cn == "" {next} {print $cn}' |
while read wday; do
echo $year/$month/$wday
done
outputs:
2021/8/4
2021/8/11
2021/8/18
2021/8/25

Answer 2

Without using cal or ncal...
#!/bin/bash
inputDate="2021/08/25"
dow=$(date -d "$inputDate" +"%a")
month=$(date -d "$inputDate" +"%m")
outputDates=""
for x in $(seq 0 9)
do
validDate=$(date -d "$x $dow 5 week ago" +"%Y/%m/%d" | grep "/$month/")
if [ ! -z $validDate ]
then
if [ ! -z $outputDates ]
then
outputDates="$outputDates,$validDate"
else
outputDates="$validDate"
fi
fi
done
echo "$outputDates"
This script outputs:
2021/08/04,2021/08/11,2021/08/18,2021/08/25