Left padding in Unix

Shell Programming and Scripting
1

I am passing input string,length, and the pad character.

input string=123
Pad char=#
Length=6

then the output should be:

###123

How we can do this?

Thanks

2

awk, perl, c, or shell (if shell, which shell?) ?

3

AWK or KSH.

4 
#!/bin/ksh
lpad()
{
   if [ $# -eq 3 ] ; then
     echo "$1" | awk -v pad="$2" -v len=$3 '{
        if(length($0) >= len) {print $0; exit}
        for(i=0;i<len - length($0);i++) { tmp=tmp pad }
        print tmp $0 }'
   else
     echo "Invalid parms for lpad" > 2   
   fi
}

# usage / test
val=$(lpad "test string" "0" 14); echo "$val"
val=$(lpad 1234 "#"  4);  echo "$val"
val=$(lpad 1234 "#"  8);  echo "$val"
5

Here is a solution using pure ksh - it's much bigger that the version that spawns awk but you do get substr() and fill() functions at no extra cost:

substr() {
  # string start [length]
  str="$1"
  start=$(( $2 ))
  let end=${#str}-$start-$(( ${3:-9999} ))+1
  [ $start -gt ${#str} ] && return
  for chop in "??????????" "????" "?"
  do
    while [ $start -gt ${#chop} ]
    do
      str="${str#$chop}"
      let start=start-${#chop}
    done
    while [ $end -ge ${#chop} ]
    do
        str="${str%$chop}"
        let end=end-${#chop}
    done
  done
  printf "$str"
}
fill() {
    # fill count [chars]
    FILL="${2:- }"
    chars=$(( $1 ))
    c=0
    while [ $c -le $chars ]
    do
        printf "%s" "$(substr "$FILL" 1 ${chars}-$c )"
        let c=c+${#FILL}
    done
}
lpad() {
  # lpad width string [chars]
  let PAD=$1-${#2}
  if [ $PAD -lt 1 ]
  then
    substr "$2" 1 $1-1
  else
    fill $PAD "$3"
    printf %s "$2"
  fi
}
 
#and now some tests
echo $(lpad 14 "test string" ".-")
echo $(lpad 4 1234 "#")
echo $(lpad 8 1234 "#")

output:

.-.test string
123
####1234
6

@pandeesh: Try Perl.

perl -e '$str="123";$pad="#";$len=3;print (($pad x $len) . $str)'

Just a funny note. No offense intended :smiley:
Master Foo and the Ten Thousand Lines