Passing argument to AWK

Question

I have the following AWK command which works well.

awk -v RS="\n+?[$]{4}\n+?" '/HMDB0000008/' test.xt

However what I want is to be able to call it from the command line and pass an argument so that it can find what I want.

For example, something like this:

awk -f var=HMDB0000008 RS="\n+?[$]{4}\n+?" '/$var/' test.txt

However this does not seem to work. Can you tell me what would be the best way?

Thank you

Does this answer your question? Pass shell variable as a /pattern/ to awk — thanasisp, Nov 25 '20 at 18:13
Why get an answer on SO and then immediately ask a question about that answer here on SE? — Ed Morton, Nov 25 '20 at 20:28

jesse_b · Accepted Answer · 2020-11-25T18:12:50.620

5

awk doesn't use $ to expand variables, additionally they are passed with the -v option not -f. Also strings within /.../ will be treated as a regex pattern so you need to use the ~ pattern match operator against the entire line ($0):

awk -v var=HMDB0000008 -v RS="\n+?[$]{4}\n+?" '$0 ~ var' test.txt

edited Nov 25 '20 at 18:12

answered Nov 25 '20 at 17:49

jesse_b

37,005

2

Also awk regexps don't have perl's +? operator. Even if it was supported, it wouldn't make much sense here, the first one would make no difference compared the + and the second is superflous as a non-greedy \n+? followed by nothing is the same as just \n. (using a regexp as RS is also not standard/portable) – Stéphane Chazelas Nov 25 '20 at 19:19
1

See also index() to find substrings. – Stéphane Chazelas Nov 25 '20 at 19:20

Passing argument to AWK

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