Bash – How to loop through dates using Bash


I have such bash script:

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

for i in "${array[@]}"
    python /home/user/ {i} &> /home/user/${i}.log

Now I want to loop through a range of dates, e.g. 2015-01-01 until 2015-01-31.

How to achieve in Bash?


Nice-to-have: No job should be started before a previous run has completed. In this case, when is completed bash prompt $ will return.

e.g. could I incorporate wait%1 in my loop?

Best Answer

Using GNU date:

while [ "$d" != 2015-02-20 ]; do 
  echo $d
  d=$(date -I -d "$d + 1 day")

  # mac option for d decl (the +1d is equivalent to + 1 day)
  # d=$(date -j -v +1d -f "%Y-%m-%d" $d +%Y-%m-%d)

Note that because this uses string comparison, it requires full ISO 8601 notation of the edge dates (do not remove leading zeros). To check for valid input data and coerce it to a valid form if possible, you can use date as well:

# slightly malformed input data

# After this, startdate and enddate will be valid ISO 8601 dates,
# or the script will have aborted when it encountered unparseable data
# such as input_end=abcd
startdate=$(date -I -d "$input_start") || exit -1
enddate=$(date -I -d "$input_end")     || exit -1

while [ "$d" != "$enddate" ]; do 
  echo $d
  d=$(date -I -d "$d + 1 day")

One final addition: To check that $startdate is before $enddate, if you only expect dates between the years 1000 and 9999, you can simply use string comparison like this:

while [[ "$d" < "$enddate" ]]; do

To be on the very safe side beyond the year 10000, when lexicographical comparison breaks down, use

while [ "$(date -d "$d" +%Y%m%d)" -lt "$(date -d "$enddate" +%Y%m%d)" ]; do

The expression $(date -d "$d" +%Y%m%d) converts $d to a numerical form, i.e., 2015-02-23 becomes 20150223, and the idea is that dates in this form can be compared numerically.