RegExp return the text inside quotes

I'm looking for a way to return only what is between the text in a column or a period if it is between two sets of quotes. There could be any number of occurrences in the text of the column.

"ENC". "CHECKOUT_TIME" [DATE, 7], "ENC" "." " PAT_ENC_CSN_ID "[NO.22],"ENC"". "" CHECKIN_TIME "[DATE, 7].

Becomes

"ENC". "" CHECKOUT_TIME ""ENC ". "" PAT_ENC_CSN_ID ""ENC ". "" CHECKIN_TIME ".

It's possible with a regular expression?

Hello

I don't know if this can be done with a single regular expression.

Here's a way that uses two functions:

SELECT str

REGEXP_REPLACE (REGEXP_SUBSTR (str

, '".*"'

)

, '("[^"]*"(\." [^"]*")?) [^"]*'

, '\1'

) AS new_str

FROM table_x

;

REGEXP_SUBSTR eliminates materials without quotation marks before the first quotation mark and after the last quotation mark double.

Tags: Database

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