Output desired text only

Question

I have a file as:

$ cat file
.
.
4h
5h
6h
7h
.
.
.
16h
17h
18h
.
.

I want to output only the lines that have '6h' and '7h' but using below awk it selects also '16h' and '17h' :(

awk '/6h/ || /7h/{print}' file

How can I make it select only '6h' and '7h' without the '16h' and '17h'. AWK is not mandatory if any other command does the job more easly.

http://stackoverflow.com/a/4800246/171318 – hek2mgl Mar 07 '17 at 12:49 — hek2mgl, Mar 07 '17 at 12:49

score 3 · Answer 1 · answered Mar 07 '17 at 12:42

3

How about this:

grep '^[67]h' file

answered Mar 07 '17 at 12:42

Alfe

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score -1 · Accepted Answer · edited Jun 20 '20 at 09:12

-1

With grep exact match operator:

grep -w "6h\|7h" file

From grep manual:

-w, --word-regexp

Select only those lines containing matches that form whole words. The test is that the matching substring must either be at the beginning of the line, or preceded by a non-word constituent character. Similarly, it must be either at the end of the line or followed by a non-word constituent character. Word-constituent characters are letters, digits, and the underscore.

edited Jun 20 '20 at 09:12

Community

1
1

answered Mar 07 '17 at 12:50

Zumo de Vidrio

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That would also match lines containing `123 7h`, for example – hek2mgl Mar 07 '17 at 15:10

Output desired text only

2 Answers2