CODEWITHSUNDEEP
regexsedawk
rsharma
rsharma

Reputation: 642

Numbering the occurrences themseleves using sed

How can I number the occurrences of the matched pattern using sed. For eg: I have a file with data in the format :

line1
line2
line3
********
line4
line5
********
line6
line7
line8
line9
*******
line10

And I need the output in the format below:

line1
line2
line3
1*******
line4
line5
2*******
line6
line7
line8
line9
3******
line10

So, the basic requirement is replacing the first occurrance of /^\*/ with 1, second occurrance with 2, third with 3 and so on.

Upvotes: 0

Views: 114

Answers (5)

potong
potong

Reputation: 58488

This might work for you (GNU sed):

sed -r '/^\*/!b;x;:a;s/9(_*)$/_\1/;ta;s/^(_*)$/0\1/;s/$/:9876543210/;s/(.)(_*):.*(.)\1.*/\3\2/;s/_/0/g;x;G;s/(.*)\n(.*)/\2\1/' file

As a solution it is little more than a joke, however where there's a will there's a way.

Upvotes: 2

Paul Evans
Paul Evans

Reputation: 27577

Use awk instead:

 awk '{if ($0 ~ /[*]+/) printf "%d%s\n", ++a, substr($0, 2); else print $0;}' file

Upvotes: 0

choroba
choroba

Reputation: 242028

sed cannot do any arithmetics. Use a different tool, e.g. Perl:

perl -pe 'print ++$i if /^\*+$/' file

Upvotes: 2

anubhava
anubhava

Reputation: 785651

Use the right tool for the right job. This is tailor made job for awk:

awk '$1 ~ /\*\*\*\*/{print ++a $0; next} {print}' file
line1
line2
line3
1********
line4
line5
2********
line6
line7
line8
line9
3*******
line10

Or else pure BASH solution:

while read l; do
    [[ $l == *'****'* ]] && l=$((++x))"$l"
    echo "$l"
done < file

Upvotes: 2

Kent
Kent

Reputation: 195209

try this one-liner (awk):

 awk '/^[*]+$/{$0=++i$0}7' file

Upvotes: 1

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