Question

我正在尝试编写一个mysql查询来选择表的所有行，其中一对一关系列是一对多关系的0。基本上，我希望所有已完成的线索不是1。

我的查询失败

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`

此操作失败并仍然返回所有潜在客户，即使他们的代码已完成1。

预期输出

带有id 12的引线和带有id 2的引线。

表格

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

表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  |

表call_result_codes

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

Answer 1

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)

我假设你的预期输出你想要那些在call_results_codes中没有记录的人，所以它足以让他们没有一个记录在那里，结果= 1。

Answer 2

试试这个。您在连接条件中有完成列。我放在哪里找到结果

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`

SQL其中列不是一对多关系中的一对一的SQL

2 个答案: