SQL 其中列不是一对多关系中的值

Question

我正在尝试编写一个 mysql 查询 select table 的所有行，其中一对一关系列是一对多关系的 0。基本上，我希望所有完成的线索都不是 1。

我的失败查询

SELECT * from `leads`
LEFT JOIN `call_lead` on `leads`.`id` = `call_lead`.`lead_id` 
LEFT JOIN `call_result_codes` ON `call_lead`.`call_result_code_id` = `call_result_codes`.`id` 
AND `call_result_codes`.`finished` in (0) group by `leads`.`id`

这失败了，但仍然 returns 所有潜在客户，即使他们的代码已完成 1。

预期输出为

ID 为 12 的领导和 ID 为 2 的领导。

Table领先

_____________________
|id  | name          |
|2   | test name     |                
|8   | test name2    |
|12  | test name2    |

Table call_lead

_____________________________________________________
|id | lead_id | user_id | call_result_code_id |remark|
|22 | 8       | 1       | 0                   |test  |
|23 | 8       | 1       | 1                   |test  |
|24 | 2       | 1       | 0                   |test  |

Table call_result_codes

________________________________
|id | description   | finished | 
|0  | not answering | 0        |                 
|1  | not interested| 1        |

Answer 1

试试这个。您在连接条件中有完成列。我把在哪里找到结果

SELECT * from `leads`
LEFT JOIN `call_lead` on `leads`.`id` = `call_lead`.`lead_id` 
LEFT JOIN `call_result_codes` ON `call_lead`.`call_result_code_id` = `call_result_codes`.`id` 
WHERE `call_result_codes`.`finished` = 0
group by `leads`.`id`

Answer 2

您可以使用 EXISTS() :

SELECT * FROM `leads`
LEFT OUTER JOIN `call_lead` on `leads`.`id` = `call_lead`.`lead_id` 
WHERE NOT EXISTS(SELECT 1 FROM `call_result_codes`
                 WHERE `call_lead`.`call_result_code_id` = `call_result_codes`.`id` 
                   AND `call_result_codes`.`finished` = 1)

根据您的预期输出，我假设您也想要那些在 call_results_codes 中没有记录的人，所以他们没有 finished = 1.[=12 的记录就足够了=]

SQL 其中列不是一对多关系中的值

SQL where column is not value in a one to one of a one to many relation

mysql

left-join