mysql - MySQL-CAST json将数据提取为时间

从JSON类型的数据提取时，有什么方法可以比较小时？

目前，我有此查询：

SELECT * FROM
(SELECT clinics.id as cid, clinics.lat, clinics.lng, clinics.opening_hours,
clinics.logo, clinics.address, clinics.city, clinics.state, clinics.name, clinics.description,
clinics.zip, clinics.phone_number, clinics.email, clinics.url, clinics.bookmark_url,
clinics.marker, timezone,
countries.full_name AS country,
(6378 * acos(
cos(radians(-33.8688197)) * cos(radians(lat)) *
cos(radians(lng) - radians(151.2092955)) +
sin(radians(-33.8688197)) * sin(radians(lat))))
AS distance
FROM clinics
JOIN countries ON countries.id = clinics.country_id
LEFT JOIN clinics_services ON clinics.id = clinics_services.clinic_id
WHERE CAST(JSON_EXTRACT(`opening_hours`, '$."wednesday-to"') as time) > CAST('09:00' as time)
GROUP BY clinics.id
) AS distances
WHERE distance < 5
ORDER BY distance ASC

如果运行此查询，我将得到正确的结果：

SELECT JSON_EXTRACT(`opening_hours`, '$."wednesday-to"') FROM clinics

（结果）：

"17:30"
"19:00"
"12:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"19:00"
"18:00"
"19:00"
"18:00"
"18:00"
"18:00"
"18:00"
"18:00"
"19:00"

但是，如果我尝试将其强制转换为时间（使用此查询），则会得到NULL：

SELECT CAST(JSON_EXTRACT(`opening_hours`, '$."wednesday-to"') as TIME) FROM clinics

我究竟做错了什么？

最佳答案

看来@BillKarwin是正确的。您需要用JSON_EXTRACT包装JSON_UNIQUOTE。

Chaining JSON_EXTRACT with CAST or STR_TO_DATE fails

https://www.db-fiddle.com/f/nomb47TQ2Rxaw5MGc1WtSs/0

SELECT CAST(JSON_UNQUOTE(JSON_EXTRACT('{"hour":"10:00"}','$.hour')) as time) > CAST('09:00' as time);

在您的情况下：

...
WHERE CAST(JSON_UNQUOTE(JSON_EXTRACT(`opening_hours`, '$."wednesday-to"')) as time) > CAST('09:00' as time)
...