Sed replace at second occurrence

Question

I want to remove a pattern with sed, only at second occurence. Here is what I want, remove a pattern but on second occurrence.

What's in the file.csv:

a,Name(null)abc.csv,c,d,Name(null)abc.csv,f
a,Name(null)acb.csv,c,d,Name(null)acb.csv,f
a,Name(null)cba.csv,c,d,Name(null)cba.csv,f

Output wanted:

a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

This is what i tried:

sed -r 's/(\(null)\).*csv//' file.csv

The problem here is that the regex is too greedy, but i cannot make is stop. I also tried this, to skip the first occurrence of "null":

sed -r '0,/null/! s/(\(null)\).*csv//' file.csv

Also tried but the greedy regex is still the problem.

sed -r 's/(\(null)\).*csv//2' file.csv

I've read that ? can make the regex "lazy", but I cannot make it workout.

sed -r 's/(\(null)\).*?csv//' file.csv

Hiro

If you may have 3 or more (null)s and you still want to only remove the 2nd occurrence, I think it would be easier to do with perl, using .*? instead of .*. – Mildred

Answer 1

The more robust awk solution:

Extended sample file input.csv:

12,Name(null)randomstuff.csv,2,3,Name(null)randomstuff.csv, false,Name(null)randomstuff.csv
12,Name(null)AotherRandomStuff.csv,2,3,Name(null)AotherRandomStuff.csv, false,Name(null)randomstuff.csv
12,Name(null)alphaNumRandom.csv,2,3,Name(null)alphaNumRandom.csv, false,Name(null)randomstuff.csv

---

The job:

awk -F, '{ c=0; for(i=1;i<=NF;i++) if($i~/\(null\)/ && c++==1) sub(/\(null\).*/,"",$i) }1' OFS=',' input.csv

The output:

12,Name(null)randomstuff.csv,2,3,Name, false,Name(null)randomstuff.csv
12,Name(null)AotherRandomStuff.csv,2,3,Name, false,Name(null)randomstuff.csv
12,Name(null)alphaNumRandom.csv,2,3,Name, false,Name(null)randomstuff.csv

Bonilla

Great this is Working just fine! I'll have to check out more on awk tool! – Hiro

Answer 2

sed does provide an easy way to specify which match to be replaced. Just add the number after delimiters

$ sed 's/(null)[^.]*\.csv//2' ip.csv
a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

$ # or [^,] if there are no , within fields
$ sed 's/(null)[^,]*//2' ip.csv
a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

Also, no need to escape () when not using extended regular expressions

Minneapolis

I've tried it if you look closer in my post. The problem was the Greedy Regex. I had to change .* with [^,]* like in your example. Thank you. – Hiro

well I didn't notice that you had tried //1 (later edited to //2) ... so you were only put off by greedy issue... easy to solve in this case as there are workarounds with [^,] or [^.]... for generic case you might need proper csv parsers available in perl/python/etc – Minneapolis

You are right, i could of done this with pyexcel which i use in my script. Didn't thought about that! – Hiro

ahhh, this is exactly what I needed as well, Thanks! – Bronchia

Answer 3

Execute:

awk '{sub(/.null.....csv,f/,",f")}1' file

And the output should be:

a,Name(null)abc.csv,c,d,Name,f
a,Name(null)acb.csv,c,d,Name,f
a,Name(null)cba.csv,c,d,Name,f

Ligurian