Bash and awk - how to pass a variable to awk when doing multiline pattern maths?

Question

I'm using AWK for multiline matching with patterns. Example:

awk '/"ip": /,/"id": /'

I would like to pass an argument to AWK when I know the ip, so that it would return the lines that include the specified IP. Example:

awk '/"ip": "$IP"/,/"id": /'

What would be a way to do it?

I have tried:

awk '/"ip": "$IP"/,/"id": /'
awk '/"ip": "${IP}"/,/"id": /'
awk "/\"ip\": \"$IP\"/,/\"id'\": /"

I was wondering how to use this approach with patterns:

var="hello"; awk -v a="$var"

Thanks.

ANSWER

accepted and can be found below.

Additional details for anyone interested on how to make the matching more precise (using the variable + a string pattern). Take note that you can't wrap the variable you are using in the pattern with double quotes, it will not work even if the quotes are escapted:

cat l.txt | awk '$0 ~ "\"ip\": \"103.6.181.43", $0 ~ "\"id\": \""'              ===> works
cat l.txt | awk -v ipvar=103.6.181.43 '$0 ~ ipvar, $0 ~ "\"id\": \""'           ===> works
cat l.txt | awk -v ipvar=103.6.181.43 '$0 ~ "\"ip\": \"ipvar", $0 ~ "\"id\": \""' ===> does not work
cat l.txt | awk -v ipvar=103.6.181.43 '$0 ~ "\"ip\": \""ipvar, $0 ~ "\"id\": \""' ===> works

Finally, you can wrap / surround the parameter used in qoutes. Compare:

awk -v ipvar=103.6.181.43 '$0 ~ "\"ip\": \"ipvar\"", $0 ~ "\"id\": \""'  ===> will not work
awk -v ipvar=103.6.181.43 '$0 ~ "\"ip\": \""ipvar"\"", $0 ~ "\"id\": \""' ===> will work and is equivalent.

Please add sample input and expected output. – Thor Apr 14 '16 at 15:30 — Thor, Apr 14 '16 at 15:30

score 2 · Accepted Answer · answered Apr 14 '16 at 12:35

2

Do not add the variable within the / /, because what goes there is a regular expression.

Instead, do use $0 ~ pattern, which gives you more freedom. After all,

/pattern1/,/pattern2

Is the same as

$0 ~ pattern1, $0 ~ pattern2

So you can use -v to provide the value and say:

awk -v var="$bash_variable" '$0~"ip: " var,$0 ~ "id: " var'
#                            ^^^^^^^^^^^^^
#                                          ^^^^^^^^^^^^^^^
#                               start           end

Test

$ cat a
hello
ip: 23
bye
and this is
id: 23
hehe
done
$ awk -v var=23 '$0~"ip: " var,/id: /' a
ip: 23
bye
and this is
id: 23

answered Apr 14 '16 at 12:35

fedorqui

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Thanks, will also add some more points in the OP for reference to anyone who is lost and stumbles upon the question :) – Mindaugas Bernatavičius Apr 14 '16 at 16:23
@MindaugasBernatavičius nice to read it worked to you. I kept my answer very simple. I guess your final solution is a bit more cumbersome because of all these double quotes that need to be escaped. Good thing you show the final version for future readers! – fedorqui Apr 15 '16 at 08:00

Rany Albeg Wein · Answer 2 · 2016-04-14T12:32:05.320

1

You're trying to expand a variable inside single quotes. Single quotes (') cause everything between them to be taken literally by bash. Try to use double-quotes instead, and see Quotes.

Regarding your question about awk assignments, here's an exapmle:

awk -v foo="bar" 'foo ~ $0' <<< $'foo\nbar\nbaz'

Output: bar.

edited Apr 14 '16 at 12:32

answered Apr 14 '16 at 12:21

Rany Albeg Wein

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I have tried the double qoutes - it's in the examples. Also, I don't understand how does the example with the assignment uses patterns - I would still like to preserve the multi-line match somehow :) – Mindaugas Bernatavičius Apr 14 '16 at 12:31
1

Are you trying to match `IP` sorunded by *literal* double-quotes? – Rany Albeg Wein Apr 14 '16 at 12:35
Yes, the data example is like this `"ip": "127.0.0.1",` .. then a couple of lines (count of the lines is unknown, hence I need patterns) and then `"id": "123",` – Mindaugas Bernatavičius Apr 14 '16 at 12:42
1

I see. Check out @fedorqui answer. It is better than mine. – Rany Albeg Wein Apr 14 '16 at 12:48

Bash and awk - how to pass a variable to awk when doing multiline pattern maths?

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