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Help with awk script and update position

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# 1  
Old 02-15-2012
Help with awk script and update position
Hi, I'm using AIX 5.3 and am trying to get this to work as I need:

Code:
echo 6078:  6072  6073  6074  6075  6076  6077  6078  6079 | awk '{n=$8} n{sub(n,n+1000000)}1'
1006078: 6072 6073 6074 6075 6076 6077 6078 6079

I only want it to change second occurance of 6078 (ie $8) and not the first ($1).
I tried:

Code:
echo 6078:  6072  6073  6074  6075  6076  6077  6078  6079 | awk '{n=" "$8" "} n{sub(n," "n+1000000" ")}1'
6078: 6072 6073 6074 6075 6076 6077 1006078 6079

but I'm running piping through 10 awks (one to change each number), but on the rows with less than 10, it goes haywire if I have the " " in.
# 2  
Old 02-15-2012
Perl, ok?
Code:
echo 6078:  6072  6073  6074  6075  6076  6077  6078  6079 | perl -lane '$F[7]+=1000000;print join (" ",@F)'

# 3  
Old 02-15-2012
no, because as I said, I need to call it 10 times (as some lines have 10 entries, but some may have one). With the above, if I add
Code:
perl -lane '$F[9]+=1000000;print join (" ",@F)'

when there are only 8 entries, it outputs 1000000 as the last one instead of blank. Awk doesn't do that.
# 4  
Old 02-16-2012
Code:
echo 6075:  6072  6073  6074  6075  6076  6077  6078  6079 | \
perl -e 'chomp($x=<>);@x=split/\s+/,$x;for($i=1; $i<=$#x; $i++){if($x[0]=~/$x[$i]/){$x[$i]+=1000000;last}};print join(" ", @x);'

This User Gave Thanks to balajesuri For This Post:
amrit1987
# 5  
Old 02-16-2012
Try:
Code:
awk '{p=$1;for(i=2;i<=NF;i++)p=p OFS $i+1000000;print p}'

Code:
$ echo "6078:  6072  6073  6074  6075  6076  6077  6078  6079" | awk '{p=$1;for(i=2;i<=NF;i++)p=p OFS $i+1000000;print p}'
6078: 1006072 1006073 1006074 1006075 1006076 1006077 1006078 1006079

These 2 Users Gave Thanks to Scrutinizer For This Post:
amrit1987 say170
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