Displaying Number with printf

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# 1  
Old 02-11-2008
Displaying Number with printf

I am trying to display a number with commas
printf "%d\n" 323232     
printf "%d\n" 1234567

I want the output to be:

I tried to change %d to other formats and could find the solution.
any idea?
# 2  
Old 02-11-2008
Search the Web for a script called nicenumber.sh

Aternatively use this sed one liner which handles up to 3 decimal places correctly
$ echo 1234567 | sed -e :a -e 's/\(.*[0-9]\)\([0-9]\{3\}\)/\1,\2/;ta'
$ echo 1234567.123 | sed -e :a -e 's/\(.*[0-9]\)\([0-9]\{3\}\)/\1,\2/;ta'

Last edited by fpmurphy; 02-11-2008 at 06:42 AM..
# 3  
Old 02-11-2008
works good, thanks
Do yo have a solution within awk ?
# 4  
Old 02-11-2008
well, you can do it like this with sed:

#  echo 12345678 | sed -e :x -e 's/\([0-9][0-9]*\)\([0-9][0-9][0-9]\)/\1,\2/' -e 'tx'

# 5  
Old 02-11-2008
AWK solution

I found almost working solution
echo 1 12 123 1234 12345 123456 1234567 | awk --re-interval '{print gensub(/([[:digit:]])([[:digit:]]{3})/,"\\1,\\2","g")}' 

  1 12 123 1,234 1,2345 1,23456 1,234567

any idea how to fix it ?
# 6  
Old 02-11-2008
From the gawk manual

A single quote or apostrohe character is a POSIX extension to ISO C. It indicates that the integer part of a floating point value, or the entire part of an integer decimal value, should have a thousands-separator character in it. This only works in locales that support such characters.

     $ cat thousands.awk                                   
     BEGIN { printf "%'d\n", 1234567 }
     $ LC_ALL=C gawk -f thousands.awk                          
     $ LC_ALL=en_US.UTF-8 gawk -f thousands.awk             

# 7  
Old 02-12-2008
this is not working under linux
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