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* How to do thousands grouping in zsh?
@ 2014-09-04  8:52 Han Pingtian
  2014-09-04  9:44 ` Peter Stephenson
  0 siblings, 1 reply; 2+ messages in thread
From: Han Pingtian @ 2014-09-04  8:52 UTC (permalink / raw)
  To: zsh-users

Hello,

Looks like builtin printf doesn't support "'" flag. I try to figure out
a method to do the thousands grouping of numbers. It works fine. 

	num_group()
	{
	    local a
	    local i
	
	    a=(${(s::)1})
	    i=$((${#a}/3))
	
	    for ((;i>0;i--))
	    do
	        if [[ -n $a[-i*3-1] ]]
	        then
	            a[-i*3]=(, $a[-i*3])
	        fi
	    done
	
	    print ${(j::)a}
	}

Is there any other method?

Thanks in advance! 


^ permalink raw reply	[flat|nested] 2+ messages in thread
* Re: How to do thousands grouping in zsh?
  2014-09-04  8:52 How to do thousands grouping in zsh? Han Pingtian
@ 2014-09-04  9:44 ` Peter Stephenson
  0 siblings, 0 replies; 2+ messages in thread
From: Peter Stephenson @ 2014-09-04  9:44 UTC (permalink / raw)
  To: zsh-users

On Thu, 04 Sep 2014 16:52:23 +0800
Han Pingtian <hanpt@linux.vnet.ibm.com> wrote:
> Is there any other method?

From 5.0.6, just released.

print $(( [#_3] 299792458 ))

299_792_458

It's always "_" because it's in the same format that's understood on
input.  You could postprocess it

print ${$(( [#_3] 299792458 ))//_/,}

pws


^ permalink raw reply	[flat|nested] 2+ messages in thread
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2014-09-04  8:52 How to do thousands grouping in zsh? Han Pingtian
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