grep command in unix for getting particular string

Question

I have test.txt file, which contains the lines below:

Period 1
Period 2
Period 3
........
.
.
.
Period 10
Period 11
Period 12
.
.
.

Likewise I have 18 rows inside the test.txt file.

Now I need to create the new file from this test.txt such that when I give 'Period 1', it should show only "Period 1" row alone in new file.

I am facing issue here: when I'm trying to do grep "Period 1" > newfile.txt, it gives all the period which start with "Period 1", "Period 10", "Period 11", and so on.

Or use a regex with grep. Match the end of line character if applicable. — user201891
– user201891, Commented Apr 7, 2015 at 20:10

Dmitry Ginzburg · Accepted Answer · 2015-04-07 19:52:12Z

1

You should use grep with general expression, which should match the beginning and the end of the line:

grep -e '^Period 1$' > newfile.txt

The regular expression here means: "<the beginning of the line>Period<space>1<the end of the line>"

If you're not sure about format, you may use the expression, which will match any number of spaces (or any other whitespaces) in the middle:

egrep '^Period\s*1$' # egrep is synonym for `grep -E`, where -E stands for extended regexps

For further reading on regular expressions in grep: https://www.digitalocean.com/community/tutorials/using-grep-regular-expressions-to-search-for-text-patterns-in-linux

edited Apr 7, 2015 at 19:52

answered Apr 7, 2015 at 19:47

Dmitry Ginzburg

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