Regex to get everything between first and last double quote but without double quote

Question

I have following data

"full name":"vg","address":"abc, 128, street no 9,"(0112093-0)""

Now it is clear in "key":"value" format and it is separated by comma.So I have 2 keys "full name" , "address" . I need to write a regex which will get sting between double quotes only , so something like

full name : vg
address : abc, 128, street no 9,"(0112093-0)"

I tried following but I am getting double quotes along

"(\w+)":"(.*?)"

I want to remove the fist and last double quotes from both key and value.

you mean double quotes within address value ? Actually it is fine for me if it comes anywhere in between since whole sting is between first and last double quotes — user3332404, Aug 18 '17 at 16:32
I am sorry I am new to regex so I am just taking help from google — user3332404, Aug 18 '17 at 16:34
I also tried following but i am getting both values and keys with double quotes ((?:"[^"]*"|[^:,])*):((?:"[^"]*"|[^,])*) — user3332404, Aug 18 '17 at 16:42
There is no solution since data is badly formatted. There is no difference between inner quotes and closing quotes. — anubhava, Aug 18 '17 at 16:53
@anubhava That is not true - key is quoted string without followed by color followed by quoted string. Seems straight forward. — NetMage, Aug 18 '17 at 17:27
@user3332404 What language/environment? How are you removing outside quotes - replace? — NetMage, Aug 18 '17 at 17:27
`Seems straight forward` Look at the value part of `address` field :) — anubhava, Aug 18 '17 at 17:38

score 0 · Answer 1 · answered Aug 19 '17 at 04:48

0

You can use /\"(\w+)":"(.*?)"(?=,|$)/g. Test it online

I have assumed a value can terminate with ", (might be a boundary case) or "$.

answered Aug 19 '17 at 04:48

abskmj

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Regex to get everything between first and last double quote but without double quote

1 Answers1