How to loop through dates using Bash?

Question

I have such bash script:

array=( \'2015-01-01\', \'2015-01-02\' )

for i in \"${array[@]}\"
do
    python /home/user/executeJobs.py {i} &> /home/user/         


        

Answer 1

I had the same issue and I tried some of the above answers, maybe they are ok, but none of those answers fixed on what I was trying to do, using macOS.

I was trying to iterate over dates in the past, and the following is what worked for me:

#!/bin/bash

# Get the machine date
newDate=$(date '+%m-%d-%y')

# Set a counter variable
counter=1 

# Increase the counter to get back in time
while [ "$newDate" != 06-01-18 ]; do
  echo $newDate
  newDate=$(date -v -${counter}d '+%m-%d-%y')
  counter=$((counter + 1))
done

Hope it helps.

Answer 2

Brace expansion:

for i in 2015-01-{01..31} …

More:

for i in 2015-02-{01..28} 2015-{04,06,09,11}-{01..30} 2015-{01,03,05,07,08,10,12}-{01..31} …

Proof:

$ echo 2015-02-{01..28} 2015-{04,06,09,11}-{01..30} 2015-{01,03,05,07,08,10,12}-{01..31} | wc -w
 365

Compact/nested:

$ echo 2015-{02-{01..28},{04,06,09,11}-{01..30},{01,03,05,07,08,10,12}-{01..31}} | wc -w
 365

Ordered, if it matters:

$ x=( $(printf '%s\n' 2015-{02-{01..28},{04,06,09,11}-{01..30},{01,03,05,07,08,10,12}-{01..31}} | sort) )
$ echo "${#x[@]}"
365

Since it's unordered, you can just tack leap years on:

$ echo {2015..2030}-{02-{01..28},{04,06,09,11}-{01..30},{01,03,05,07,08,10,12}-{01..31}} {2016..2028..4}-02-29 | wc -w
5844