Question

我刚刚在数据库中创建了4个新列，分别名为cea_no，district，property_type和listing_type。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;

采样结果正确

+------------------------------+----------+------------------------+--------------+
| 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"    |
+------------------------------+----------+------------------------+--------------+

那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;

Answer 1

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id值匹配的适当条件下加入。这应该起作用：

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{{1}}部分中使用SET公式来编写代码：

UPDATE

Answer 2

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。就像这样

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;

根据json选择查询更新多个表列值

2 个答案: