我刚刚在数据库中创建了名为cea_no
,district
,property_type
和listing_type
的4个新列。我想将基于选择查询的结果插入到我添加的新列中。选择查询结果来自行json
,并从json
数据中提取。我该如何实现?我尝试了一些方法,它奏效了,问题是它插入了新行,现在我的数据增加了一倍。
我的表格结构。
+------------------+------------+------+-----+---------------------+-------------------------------+
| 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() |
+------------------+------------+------+-----+---------------------+-------------------------------+
我尝试过的查询
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" |
+------------------------------+----------+------------------------+--------------+
这是我要在新列中插入的值。
编辑:我尝试了此查询,但无法正常工作
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;
您需要使用INNER JOIN
,而不是CROSS JOIN
,否则将插入不正确的数据。并且您需要在适当的条件下加入,即id
值匹配。这应该工作:
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JSON_EXTRACT
的SET
部分中使用UPDATE
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)
实际上,您的查询是正确的。您只需要在子查询中添加别名即可使用。像这样
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) as x
set cea_no = cea_no,
district = district,
property_type = property_type,
listing_type = listing_type;