我刚刚在我的数据库中创建了 4 个新列,分别命名为 cea_no
、district
、property_type
和 listing_type
。我想根据我的选择查询将它们的结果插入到我添加的新列中。选择查询结果来自行 json
,它是从 json
数据中提取的。我怎样才能做到这一点?我尝试了一些方法并且有效,问题是它插入了一个新行并且我的数据现在加倍了。
我的表结构。
+------------------+------------+------+-----+---------------------+-------------------------------+
| Field | Type | Null | Key | Default | Extra |
+------------------+------------+------+-----+---------------------+-------------------------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| json | mediumtext | NO | | NULL | |
| property_name | text | NO | | NULL | |
| property_address | text | NO | | NULL | |
| price | text | NO | | NULL | |
| listed_by | text | NO | | NULL | |
| contact | text | NO | | NULL | |
| cea_no | text | NO | | NULL | EMPTY for now |
| district | text | NO | | NULL | EMPTY for now |
| property_type | text | NO | | NULL | EMPTY for now |
| listing_type | text | NO | | NULL | EMPTY for now |
| update_time | timestamp | NO | | current_timestamp() | on update current_timestamp() |
+------------------+------------+------+-----+---------------------+-------------------------------+
我试过的查询
SELECT JSON_EXTRACT(json, '$.agencyLicense') AS cea_no,
JSON_EXTRACT(json, '$.district') AS district,
JSON_EXTRACT(json, '$.details."Type"') AS property_type,
RIGHT(JSON_EXTRACT(json, '$.details."Type"'),9) AS listing_type
from xp_guru_listings;
正确的样本结果
+------------------------------+----------+------------------------+--------------+
| cea_no | district | property_type | listing_type |
+------------------------------+----------+------------------------+--------------+
| "CEA: R017722B \/ L3009740K" | "(D25)" | "Apartment For Sale" | For Sale" |
| "CEA: R016023J \/ L3009793I" | "(D25)" | "Condominium For Sale" | For Sale" |
| "CEA: R011571E \/ L3002382K" | "(D25)" | "Condominium For Sale" | For Sale" |
| "CEA: R054044J \/ L3010738A" | "(D21)" | "Apartment For Sale" | For Sale" |
| "CEA: R041180B \/ L3009250K" | "(D09)" | "Condominium For Sale" | For Sale" |
+------------------------------+----------+------------------------+--------------+
这是我要插入新列的值。
编辑: 我试过这个查询,但它不起作用
update xp_guru_listings cross join (
SELECT JSON_EXTRACT(json, '$.agencyLicense') AS cea_no,
JSON_EXTRACT(json, '$.district') AS district,
JSON_EXTRACT(json, '$.details."Type"') AS property_type,
RIGHT(JSON_EXTRACT(json, '$.details."Type"'),9) AS listing_type
from xp_guru_listings
)
set cea_no = cea_no,
district = district,
property_type = property_type,
listing_type = listing_type;
最佳答案
您需要使用INNER JOIN
,而不是CROSS JOIN
,否则您将插入不正确的数据。您需要在适当的条件下加入,即 id
值匹配。这应该有效:
update xp_guru_listings x
join (
SELECT id,
JSON_EXTRACT(json, '$.agencyLicense') AS cea_no,
JSON_EXTRACT(json, '$.district') AS district,
JSON_EXTRACT(json, '$.details."Type"') AS property_type,
RIGHT(JSON_EXTRACT(json, '$.details."Type"'),9) AS listing_type
FROM xp_guru_listings) j ON j.id = x.id
set x.cea_no = j.cea_no,
x.district = j.district,
x.property_type = j.property_type,
x.listing_type = j.listing_type;
请注意,您可以直接在 UPDATE
的 SET
部分使用 JSON_EXTRACT
公式更简单地编写此代码:
UPDATE xp_guru_listings
SET cea_no = JSON_EXTRACT(json, '$.agencyLicense'),
district = JSON_EXTRACT(json, '$.district'),
property_type = JSON_EXTRACT(json, '$.details."Type"'),
listing_type = RIGHT(JSON_EXTRACT(json, '$.details."Type"'),9)
关于mysql - 基于json select查询更新多个表列值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59277919/