Adding a zero to single digit variable

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
     开发者_高级运维       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]}, where x and y are either integers or single characters, and incr, an optional increment, is an integer.

When integers are supplied, the expression expands to each number between x and y, inclusive.

Supplied integers may be prefixed with 0 to force each term to have the same width. When either x or y begins 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

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

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

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

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