Pattern match at a specific column number with a particular string . No delimiter or separator

Question

I want to search a file to output rows matching a particular value at specific column number. There is no delimiter involved. For example:

IloveA
IloveB
IloveC
IloveA
IloveD
IloveA
Ilove A

Output :

IloveA
IloveA
IloveA

Here the important point to note it should print only if A is present at 6th column.

If you're interested in a regex-based solution, see https://stackoverflow.com/questions/34789929/regex-to-match-character-in-specific-position — oguz ismail, Aug 04 '20 at 08:02

score 0 · Answer 1 · answered Aug 04 '20 at 08:49

0

Use grep '^.....A' or more generally grep -E '^.{5}A'.

answered Aug 04 '20 at 08:49

August Karlstrom

10,773
7
38
60

Ruslan Osmanov · Answer 2 · 2020-08-04T09:23:11.673

As I understand the problem, you need to print the rows matching a pattern within the range of the first 5 characters, but only if the 6^th character is A. This can be done in two steps:

filter out the rows with the 6^th character equal to A and capture the first 5 characters at the same time;
print the line, if the captured sequence matches your pattern.

The following is a sample implementation of the above in Perl:

perl -l -n -e 'print if /^(.{5})A/ and $1 =~ /ove/' input-file.txt

The Perl code prints the next line read from the file (implicit argument $_) if the line matches /^(.{5})A/. As a result of capturing, $1 (the first and the only capturing group) contains the first five characters. Then, if $1 matches /ove/ (a sample pattern that you'd change), the second if condition is true, and the line is printed to standard output.

Pattern match at a specific column number with a particular string . No delimiter or separator

2 Answers2