bash – Print dates in a range.

Problem:
Needed to have a range of dates for a script.

Solution:
Opted to write it in bash.

How it works:

shell> bash list_dates.sh "2011-07-02" "2011-07-10"
2011-07-02
2011-07-03
2011-07-04
2011-07-05
2011-07-06
2011-07-07
2011-07-08
2011-07-09
2011-07-10

Source:

#!/bin/bash
#=============================================================================
#
# FILE: list_dates.sh
#
# USAGE: list_dates.sh start_date end_date
#
# DESCRIPTION: List the dates between a given range of dates.
#
#=============================================================================

start_date="$1"
end_date="$2"

#=== FUNCTION ================================================================
#           NAME: usage
#    DESCRIPTION: Display usage information for this script.
#    PARAMETER 1: ---
#=============================================================================
function usage()
{
cat <<- EOT
Lists the dates between a given range of dates.
Dates must be in yyyy-mm-dd format

usage: $0 start_date end_date [--help]
    start_date (YYYY-MM-DD) -- First date in the range. 
    end_date (YYYY-MM-DD) -- Last date in the range. 
    help -- display this message
EOT
}

#-----------------------------------------------------------------------------
# Display usage
#-----------------------------------------------------------------------------
if [ "$1" == "--help" ]
then
    usage
    exit
fi
#-----------------------------------------------------------------------------
# Check the date formats
#-----------------------------------------------------------------------------
start_date=$(date --date="$start_date" +"%F") # Defaults to current date if none supplied.
if [ "$?" -eq "1" ] || [ -z "$start_date" ]
then
    echo "Please provide a start date in this format (YYYY-MM-DD)"
    exit
fi

end_date=$(date --date="$end_date" +"%F") # Defaults to current date if none supplied.
if [ "$?" -eq "1" ] || [ -z "$end_date" ]
then
    echo "Please provide an end date in this format (YYYY-MM-DD)"
    exit
fi

# Make sure that the first day is always before the second date.
# If a user enters 2012-09-08 2011-02-10 switch the dates to be
# 2011-02-10 2012-09-08
file_=$(mktemp)
printf "%s\n%s" $start_date $end_date | sort > $file_
start_date=$(head -n 1 $file_)
end_date=$(tail -n 1 $file_)
rm $file_
#-----------------------------------------------------------------------------
# main
#-----------------------------------------------------------------------------
i=0
while [ "$date_" != "$end_date" ]
do
    date_=$(date --date="$start_date +$i day" +"%F")
    echo $date_
    i=$(expr $i + 1)
done

Update 2012-11-06:
Changed

date_=$(date --date="$(date --date=$start_date +%F) + $i day" +"%F")

to

date_=$(date --date="$start_date +$i day" +"%F")

ssh login without a password

Problem:
ssh login without a password

Solution:
This is a common enough problem. There are a many of solutions on the net. This one worked for me.
http://maduraisenthilscorner.blogspot.com/2011/02/ssh-login-without-password.html

NB.
In my case the machine was already configured to ssh to some servers without specifying a password. I just needed to add another server to the list. So all I had to do from the instructions from Senthil Kumar’s blog was

senthil@linux-5xxh:~> cat .ssh/id_rsa.pub | ssh siva@192.168.2.20 ‘cat >> .ssh/authorized_keys’
Password:

Once I copied over the public key. I was able to ssh without a password.

Printing variables side by side – part 1

Problem:
Print variables side by side.

Solution:
printf command For the example below echo would be best. But for a better example of when you might want to use printf read this.

shell>val1="cat"
shell>val2="dog"
shell>echo "$val1,$val2"
cat, dog
shell>printf "%s, %s" $val1 $val2
cat, dog
shell>printf "The values are %s and %s." $val1 $val2
The values are cat and dog.

Cases where you would want to use printf instead of echo are explained here
http://stackoverflow.com/questions/3652524/bash-echo-string-that-starts-with

Update. (2012-04-20)

A better example of using printf

MySQL – Remove a character from a string

Problem:
Had to strip out a plus sign at the beginning of some text in the database.

Solution:
Update command and replace function in MySQL. In this case I was getting rid of the plus sign.

UPDATE `table` SET db_field=REPLACE(db_field, ‘+’, ”) where db_field like ‘+3989%’ limit 5;

Source:
http://eisabainyo.net/weblog/2007/02/26/remove-the-first-character-in-mysql/