MySQL IN只返回一行，而JOIN查询返回所需的结果？

Question

我有3张桌子，下面给出了结构

table_incident

+-------+-------------+------------+-----------------------------------+
| id(PK)    | incident_display_id | account_id | customized_fields_id  |
+-------+-------------+------------+-----------------------------------+
| 47614     |      33             |        394 | 1285,1286,1287,1288   |
+-------+-------------+------------+-----------------------------------+

table_customized_fields_data

+------+------------+----------+-------------+--------------------+
| id   | account_id | field_id(FK) | incident_id(FK) | value      |
+------+------------+----------+-------------+--------------------+
| 1285 |        394 |       49     |   47614         | Nikon 5MP  |
| 1286 |        394 |       50     |   47614         | CDMA       |
| 1287 |        394 |       51     |   47614         | Yes        |
| 1288 |        394 |       84     |   47614         | 9317001007 |
+------+------------+----------+-------------+--------------------+

table_customized_fields

+----+------------+------------+---------------------+------+-------------+
| id | account_id | field_type | label               | name | field_lable |
+----+------------+------------+---------------------+------+-------------+
| 49 |        394 | text_field | Camera              |      |             |
| 50 |        394 | checkbox   | Cellphone           |      | CDMA, GSM   |
| 51 |        394 | radio      | Sunglasses          |      | Yes, No     |
| 52 |        394 | textarea   | Credit Card         |      | 5           |
| 83 |        394 | radio      | Cowboy Hat          |      | Yes,No      |
| 84 |        394 | text_field | Emergency Contact # |      |             |
+----+------------+------------+---------------------+------+-------------+

现在我想只选择那些标签和table_incident中存在的标签价值，

我解雇了以下查询

SELECT ti.id IncID,tcf.id labelID,tcfd.id dataID,tcf.label, tcfd.value
FROM  table_customized_fields_data tcfd  
INNER JOIN table_incident ti  ON (ti.id = tcfd.incident_id)
INNER JOIN table_customized_fields tcf ON (tcf.id = tcfd.field_id)
WHERE tcfd.id IN (ti.customized_fields_id)
AND ti.id=47614


+-------+---------+--------+--------+-----------+
| IncID | labelID | dataID | label  | value     |
+-------+---------+--------+--------+-----------+
| 47614 |      49 |   1285 | Camera | Nikon 5MP |
+-------+---------+--------+--------+-----------+

但只返回一行，请大家告诉我，虽然每个子查询单独完成，但我在做什么错误。

注意：虽然此查询会重新生成所需的数据：

SELECT tcf.id AS labelID,tcf.label,tcfd.id AS dataID, tcfd.value
FROM table_customized_fields_data tcfd 
JOIN table_customized_fields tcf ON tcf.id=tcfd.field_id
JOIN table_incident ti ON ti.id=tcfd.incident_id
WHERE ti.id=47614

但我认为必须有一些优化方式，请分享您的想法。

由于

Answer 1

您的第二个查询是优化的方式。只需确保您在table_customized_fields_data.incident_id上有索引。

此外，我认为您误解了IN的工作原理，引用手册：

  

expr IN（值，...）

     

如果expr等于IN列表中的任何值，则返回1，否则返回0 ...

所以你的

tcfd.id IN (ti.customized_fields_id)

相当于：

tcfd.id = ti.customized_fields_id

Answer 2

ti.customized_fields_id是一个字符串，而不是一个数组。它不匹配，但你得到一行，因为它是一个左连接，它是mysql。

你想要的是这样的：

SELECT ti.id IncID,tcf.id labelID,tcfd.id dataID,tcf.label, tcfd.value
FROM  table_customized_fields_data tcfd  
INNER JOIN table_incident ti  ON ti.id = tcfd.incident_id 
    AND concat(',', customized_fields_id, ',') like concat('%,', tcfd.id, ',%')
INNER JOIN table_customized_fields tcf ON tcf.id = tcfd.field_id
WHERE ti.id=47614

，当用逗号分隔时，tcfd.id在字符串customized_fields_id中