Trying to add a zero before the varaible if it's less than 10 and create said directory. I can't seem to get the zero to add correctly. Keeps resulting in making 02.1.2011, 02.2.2011 etc,etc.
i=0 for i in {01..31} do if $i > 10 then mkdir $path/02.0$i.2011 else mkdir $path/02.$i.2011 fi done You can replace the whole lot with:
for day in 0{1..9} {10..31} ; do mkdir ${path}/02.${day}.2011 done while still not having to start up any external processes (other than what may be in the loop body).
That's probably not that important here since mkdir is not one of those things you tend to do a lot of in a tight loop but it will be important if you write a lot of your quick and dirty code in bash.
Process creation is expensive when you're doing it hundreds of thousands of times as some of my scripts have occasionally done :-)
Example so you can see it in action:
pax$ for day in 0{8..9} {10..11}; do echo ${day}; done 08 09 10 11 And, if you have a recent-enough version of bash, it will honor your request for leading digits:
A sequence expression takes the form
{x..y[..incr]}, wherexandyare either integers or single characters, andincr, an optional increment, is an integer.When integers are supplied, the expression expands to each number between
xandy, inclusive.Supplied integers may be prefixed with
0to force each term to have the same width. When eitherxorybegins with a zero, the shell attempts to force all generated terms to contain the same number of digits, zero-padding where necessary.
So, on my Debian 6 box, with bash version 4.1.5:
pax$ for day in {08..11} ; do echo ${day} ; done 08 09 10 11 You can use
$(printf %02d $i) To generate the numbers with the format you want.
for i in $(seq 0 1 31) do mkdir $path/02.$(printf %02d $i).2011 done 
seq 31 would do the trick, I think. OP didn't want 0 and the default for both first and increment is 1.
seq -w generates numbers with leading zeros.
echo "this is a $var" -> echo "this is a $(printf %03d $var)"Better:
for i in $(seq -f %02g 1 31) do mkdir "$path/02.$i.2011" done Or even:
for i in {01..31} do mkdir "$path/02.$(printf "%02d" $i).2011" done 
In Bash 4, brace range expansions will give you leading zeros if you ask for them:
for i in {01..31} without having to do anything else.
If you're using earlier versions of Bash (or 4, for that matter) there's no need to use an external utility such as seq:
for i in {1..31} or
for ((i=1; i<=31; i++)) with either of those:
mkdir "$path/02.$(printf '%02d' "$i").2011" You can also do:
z='0' mkdir "$path/02.${z: -${#i}}$i.2011" Using paxdiablo's suggestion, you can make all the directories at once without a loop:
mkdir "$path"/02.{0{1..9},{10..31}}.2011 
How about using equal-width option (-w) in seq command:
for i in `seq -w 1 31`; do echo $i; done It returns:
01 02 03 ... 31 Will this work for you?
zeroes="0000000" pad=$zeroes$i echo ${pad:(-2)} 
$ seq --version | head -1 seq (GNU coreutils) 8.21 $ seq -f "%02g" 1 10 01 02 03 04 05 06 07 08 09 10 Your if/then statement is backwards. You are adding a 0 when it is a above 10, not adding one when it is below.
Another bug is that you make the cutoff strictly greater than 10, which does not include 10, even though 10 is two digits.
path=/tmp ruby -rfileutils -e '1.upto(31){|x| FileUtils.mkdir "'$path'/02.%02d.2011" % x}' Here's what I did for a script where I'm asking for a day, range of days, regex, etc. with a default to the improved regexes shared by paxdiablo.
for day in $days do if [ 1 -eq "${#day}"] ; then day="0$day" fi I just run through this at the beginning of my loop where I run a bunch of log analysis on the day in question, buried in directories with leading zeroes.
I created this simple utility to perform this, Good Luck , hope this helps stack'ers!!
for i in {1..24} do charcount=`echo $i|wc -m` count=`expr $charcount - 1` if [ $count -lt 2 ]; then i="0`echo $i`" fi echo "$i" done 
