How to use regex using awk

Question

I have a file, say test.txt which looks like a json and it contains a single line of data:

{"a":"b"},{"c":"d, e, f"},{..} and so on

I need to save all the data within each pair of braces in a different file in separate lines.

For eg: result.txt having

"a":"b"
"c":"d, e, f"

I used awk with regex in bash scripting

awk '/\{(.*?)\}/' test.txt  > result.txt

But instead of cropping up individual braces, it's printing the exact test file as it is.

Can anyone say what's going wrong?

It is actually a json response to a query to extract the properties of a certain page. I saved it in the txt file to sort out the values. I know I should use a json parser but I tried doing it this way. — Kritzii
– Kritzii, Commented Jan 12, 2017 at 9:52

P.... · Accepted Answer · 2017-01-12 08:19:19Z

1

echo '{"a":"b"},{"c":"d e f"},{..}' |awk -v RS="{|}" '{gsub(/,/,"")} /./{print $0}'
"a":"b"
"c":"d e f"
..

Note: It's a good idea to use json editing tools like jq to play around with json files.

answered Jan 12, 2017 at 8:19

P....

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5 Comments

Kritzii Over a year ago

It does work on the console. But when I try to assign the result to a file, the results don't seem to appear in new lines.

P.... Over a year ago

Its working for me . try echo bla bla bla | grep -oP '{\K.*?[^}]+' as alternate solution

Inian Over a year ago

@Kritzii: That is the exactly the problem of using text processing tools which rely on regEx in one way or the other, the slightest change in JSON formatting like an extra space will screw everything up

P.... Over a year ago

In total agreement with Inian. however, I am running over old Linux box which do not have jq and cannot get it due to some management reasons (not is my control ) then people have to search for alternatives. Reiterating , getting jq is best solution if possible.

Kritzii Over a year ago

Thanks @PS , the code worked. I have used Jackson parser before. I just tried working with awk as I am new to scripting.

Maroun · Accepted Answer · 2017-01-12 08:21:59Z

0

Or, you can use grep with lookarounds to capture only the value between curly braces:

grep -oP '(?<={).*?(?=})' test.txt > result.txt

answered Jan 12, 2017 at 8:21

Maroun

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