Replacing an string on a field via MySQL - mysql

Question

I have the following field (which is a string) of one of my tables:
'[{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
{"id":"7","value":"Reuters"},
{"id":"20","value":"2017/03/17"}]'`
As you can see, the last part, the one with id=20 has a date on it, 2017/03/17.
Can someone tell me how can I update that field so I can get the output 2017-03-17?
My desired output after the update would be:
'[{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
{"id":"7","value":"Reuters"},
{"id":"20","value":"2017-03-17"}]'`
If I use:
UPDATE mytable
SET myfield = replace(myfield , '/', '-');
It removes the / of the links, and I want to mantain them.
PS: The ids of the fields are always the same (the one for the date is always 20 for example) the only thing that changes are the values.

Answer 1

you can do it with a query like this:
UPDATE mytable
SET myfield = CONCAT (
SUBSTRING_INDEX(myfield, '"id":"20","value":', 1)
, '"id":"20","value":',
, REPLACE (SUBSTRING_INDEX(myfield, '"id":"20","value":', -1),'/','-')
)
WHERE ....:
sample
mysql> SELECT
-> CONCAT (
-> SUBSTRING_INDEX('[{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
'> {"id":"7","value":"Reuters"},
'> {"id":"20","value":"2017/03/17"}]', '"id":"20","value":', 1),
-> '"id":"20","value":',
-> REPLACE (SUBSTRING_INDEX('[{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
'> {"id":"7","value":"Reuters"},
'> {"id":"20","value":"2017/03/17"}]', '"id":"20","value":', -1),'/','-')
-> ) AS result;
+------------------------------------------------------------------------------------------------------------------------------------------------+
| result |
+------------------------------------------------------------------------------------------------------------------------------------------------+
| [{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
{"id":"7","value":"Reuters"},
{"id":"20","value":"2017-03-17"}] |
+------------------------------------------------------------------------------------------------------------------------------------------------+
1 row in set (0,00 sec)
mysql>

Answer 2

Maybe there is a better way, But you can :
split the string into rows using { as separator
find the row with "ID=20"
do the replace
join all rows back together using GROUP_CONCAT
DEMO
SELECT
SUBSTRING_INDEX(SUBSTRING_INDEX(T.json, '{', numbers.n), '{', -1) name,
CASE WHEN INSTR(SUBSTRING_INDEX(SUBSTRING_INDEX(T.json, '{', numbers.n), '{', -1), '"id":"20"') > 0
THEN REPLACE(SUBSTRING_INDEX(SUBSTRING_INDEX(T.json, '{', numbers.n), '{', -1), '/', '-')
ELSE SUBSTRING_INDEX(SUBSTRING_INDEX(T.json, '{', numbers.n), '{', -1)
END as result
FROM
(select 1 n union all
select 2 union all select 3 union all
select 4 union all select 5) numbers
JOIN (
SELECT '[{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
{"id":"7","value":"Reuters"},
{"id":"20","value":"2017/03/17"}]' as json
) T
on CHAR_LENGTH(T.json)
-CHAR_LENGTH(REPLACE(T.json, '{', ''))>=numbers.n-1
order by
n
OUTPUT
OPTION 2
SELECT SUBSTRING(json, 1, pos_begin - 1 ) as side_left,
REPLACE(SUBSTRING(json, pos_begin, pos_end - pos_begin + 1), '/', '-') as side_middle,
SUBSTRING(json, pos_end + 1, LENGTH(json)) as side_end
FROM (
SELECT LOCATE('{"id":"20"', json) as pos_begin,
LOCATE('}', json, LOCATE('{"id":"20"', json)) as pos_end,
T.json
FROM (
SELECT '[{"id":"6","value":["http://www.google.com","http://www.google.com","new"]},
{"id":"7","value":"Reuters"},
{"id":"20","value":"2017/03/17"}]' as json
) T
) S
;
OUTPUT