I'm using this command to get my output in csv.

awk 'ORS="," {print $2}'

I get the output in the following format:


Here I would like to delete 'name' and everything before 'name' and get only the column names. Note that column names can have the work 'name' too, which I donot want to delete. So I am okay with only deleting the first occurrence of name. Can this be done with awk?

I have tried:

awk 'NR >4 {print $2}' | awk '{ORS=","}'

and various combinations, none of which worked.

  • 2
    you can do with sed sed -i 's/.*name,//' /path/youfile – francois P Jan 10 '18 at 19:36
  • That sed command removed all occurrences of column names with 'name' in their name. – user2441441 Jan 10 '18 at 19:46
  • just as asked all before name & name itself :) 'Here I would like to delete 'name' and everything before 'name' ' it is a logical and :) – francois P Jan 10 '18 at 20:12
  • 2
    @francoisP, the greedy .* eats everything up to the last occurrence of name, so if the input is blah,name,othername,bleh, othername is gone too, and only bleh is left. – ilkkachu Jan 10 '18 at 20:17
  • 2
    What does the input file look like? – glenn jackman Jan 10 '18 at 21:54

If a,b,c,name,col1,col2,col3, are the second fields of each line in the original, then you could do the test against name at the same time you pick out these (this still leaves the annoying final comma):

$ awk -vORS=, 'p {print $2}; $2 == "name" {p=1} ' input; echo

So, starting with what you had (awk 'ORS="," {print $2}'), we add a test variable p that tells if name has already been seen. We print the second field only if p has been set to a true value earlier, and set it to true if the second field happens to be name. With the tests in this order, the name column itself is not printed. We could also ignore empty lines in input by changing p {print $2} to $0 && p {print $2}, that is, make a truthy (nonempty) input line a condition for the print, along with p.

I assumed here that the original input looks like this:

x a
x b
x c
x name
x col1
x col2
x col3

Alternatively, starting from the comma-separated list a,b,c,name,somename,othername,col3,:

$ echo 'a,b,c,name,somename,othername,col3,' | 
    sed -e 's/.*,name,//' -e s'/,$//'

Note the commas on both sides of ,name, in the pattern, they keep the greedy .* from catching the later names that end in ...name.

  • This almost worked. I now get ",col1,col2,col3,". I need to remove the first comma after the name – user2441441 Jan 10 '18 at 19:49
  • I think the extra comma at the beginning might be due to leading whitespaces/new lines? – user2441441 Jan 10 '18 at 19:55
  • @user2441441, mhmh, well, I can't see the original data you used to produce the comma-separated list, so it's hard to say. Empty lines would cause that, though. – ilkkachu Jan 10 '18 at 19:58
  • Thanks. Do you know how to remove the extra comma? – user2441441 Jan 10 '18 at 20:00
  • Can you please explain how the command works awk -vORS=, 'p {print $2}; $2 == "name" {p=1} '? – user2441441 Jan 10 '18 at 20:07

Assuming file contents as below:

$ cat myfile

awk solution

$ awk -F',name,' '{print substr($0,index($0,$2))}' myfile

perl solution.

$ perl -pe 's/^.*?name,//' myfile

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