Oracle Regex grabbing value from key value pair in a string - sql

Question

Consider the following column row:
col
-------------------------
'{"day":"8","every":"2"}'
I am trying to get 8 from this string using regular expression to figure out the day.
so far I have:
SELECT
regexp_replace(col, '{"day":[^0-9]', '') as "day"
FROM
mytable;
This gives me:
day
---------------
8","every":"2"}
I am having trouble figuring out how to filter out the rest of the string from the first number forward. In my example I just want the number 8 for this row.

Answer 1

When you are lucky enough to use Oracle 12c Release 1 (12.1.0.2) or later, do take a look at JSON_VALUE
WITH t (s)
AS (
SELECT '{"day":"8","every":"2"}'
FROM DUAL
)
SELECT JSON_VALUE(s, '$.day' ) AS day
, JSON_VALUE(s, '$.every') AS every
FROM t;
DAY EVERY
--- -----
8 2

Answer 2

How about this?
SELECT
regexp_replace(col, '{"day":"([0-9]+).*', '\1') as "day"
FROM
mytable;

Answer 3

If you don't have access to JSON_VALUE() then I would recommend the following regex unless you always know the position of the day key in the JSON string:
SELECT REGEXP_REPLACE(col, '^.*"day":"(\d+)".*$', '\1') AS day
FROM mytable;
This will replace the entire string (assuming it matches!) with the contents of the first capturing group (enclosed in parentheses: (\d+)). \d indicates a digit 0-9. If you want to return NULL values as well, you can replace \d+ with \d*. If negative or non-numeric values are possible, then I would recommend the following:
SELECT REGEXP_REPLACE(col, '^.*"day":"([^"]*)".*$', '\1') AS day
FROM mytable;
This will return whatever characters that might be contained in the day key.
FYI, once you have the value, numeric or non-, you can convert it to a number safely by using TO_NUMBER() along with REGEXP_SUBSTR():
SELECT COALESCE( TO_NUMBER( REGEXP_SUBSTR( REGEXP_REPLACE( col, '^.*"day":"[^"]*".*$', '\1' ), '\d+' ) ), 0 ) AS day
FROM mytable;
Hope it helps.