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# 1  
Old 04-16-2009
total last digits

hi group,

How can I count total number of 5's which are continuous in the end. i.e. in the below string, the o/p should be 4

I just know to calculate total number of 5's
Code:
$ echo "95952325555" | awk -F "5" '{print NF-1}'

6

# 2  
Old 04-16-2009
Code:
# echo "95952325555" | awk -F"[^5]" '{print length($NF)}'
4
# echo "95952355555" | awk -F"[^5]" '{print length($NF)}'
5
# echo "959523554555" | awk -F"[^5]" '{print length($NF)}'
3

# 3  
Old 04-16-2009
hi,

i think perl is easy.

below code is more flexiable that whatever your last consecutive char is, it will give out the repeatation.


Code:
my $str="124334456777";
my @arr=split("",$str);
my $n=pop @arr;
my $cnt=1;
while($#arr >= 0){
if ((pop @arr) eq  $n ){
	$cnt++;
}
else{
 print $cnt;
 exit;
}
}

# 4  
Old 04-16-2009
Or:

Code:
% perl -le'print length((split/[^5]/,shift)[-1])' 95952325555  
4
% perl -le'print length((split/[^5]/,shift)[-1])' 959523255555
5
% perl -le'print length((split/[^5]/,shift)[-1])' 9595232555  
3

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