script in bash with string

Question

I have a complete list of names separated by commas:

second_surnameA first_surnameA, nameA second_surnameB first_surnameB, nameB second_surnameC first_surnameC, nameC ....

So I'm trying to capture with a bash script the name of all persons (A, B, C) in each file:

  second_surnameA first_surnameA, nameA

  second_surnameB first_surnameB, nameB

  second_surnameC first_surnameC, nameC

I executed:

cat names_file.txt | tr ',' '\n'

is almost what I need but not enouth. Any idea?

Quasímodo · Accepted Answer · 2020-03-01 18:43:49Z

1

With awk:

awk '{for (i=1;i<=NF;i=i+3) print $i,$(i+1),$(i+2)}' inputfile

It simply uses blank spaces as delimiters to each field and prints three of them for each line.

answered Mar 1, 2020 at 18:43

Quasímodo

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builder-7000 · Accepted Answer · 2020-03-01 18:50:10Z

1

With GNU sed:

sed -re 's/(\S+ +){3}/&\n/g' input_file | sed 's/ $//' > output_file

The pattern (\S+ +){3} matches three times {3} a group made of: one or more non-spaces \S+ followed by one or more spaces. The replacement string &\n appends a new line to each matched pattern. The second sed removes trailing spaces and writes to output_file

edited Mar 1, 2020 at 18:50

answered Mar 1, 2020 at 18:35

builder-7000

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