為什么在MySQL查詢中添加“ where id”會返回零結果？

Question

此查詢正確返回3個訂單，其中一個ID為12。

select * from `orders` where `is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `partner_id` = '2')

如果我只想查詢該訂單，我會在id = 12處添加，所以我這樣做：

select * from `orders` where `is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `partner_id` = '2') and `orders`.`id` = '12'

現在，它返回零結果。 我在服用瘋葯嗎？ 我的語法沒有任何問題，所有列均明確引用了正確的表，依此類推。

如果有幫助，Laravel會自動生成。 這是laravel代碼：

這將返回3個結果：

$order = Order::completed()
    ->whereHas('orderPayments', function ($query) use ($partner) {
        $query->where('partner_id', $partner->id);
    })->get();

這沒有返回：

$order = Order::completed()
    ->whereHas('orderPayments', function ($query) use ($partner) {
        $query->where('partner_id', $partner->id);
    })->where('id', 12)->first();

編輯：

為了澄清起見，此查詢正確返回了順序。

select * from `orders` where `is_completed` = '1' and `orders`.`id` = '12'

進行where id = 12返回正確的結果，與exists ，但不會一起返回。 如果單獨執行，則兩者均不返回任何結果，而均返回正確的結果。 那是沒有道理的。

編輯：訂單表描述：

+------------------------+------------------+------+-----+---------+----------------+
| Field                  | Type             | Null | Key | Default | Extra          |
+------------------------+------------------+------+-----+---------+----------------+
| id                     | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| user_id                | int(11)          | YES  | MUL | NULL    |                |
| is_completed           | tinyint(1)       | NO   |     | 0       |                |
| created_at             | timestamp        | YES  |     | NULL    |                |
| updated_at             | timestamp        | YES  |     | NULL    |                |
+------------------------+------------------+------+-----+---------+----------------+

和order_payments表：

+----------------------------+------------------+------+-----+---------+----------------+
| Field                      | Type             | Null | Key | Default | Extra          |
+----------------------------+------------------+------+-----+---------+----------------+
| id                         | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| order_id                   | int(11)          | NO   | MUL | NULL    |                |
| partner_id                 | int(11)          | NO   | MUL | NULL    |                |
| amount_in_pennies          | int(11)          | YES  |     | NULL    |                |
| created_at                 | timestamp        | YES  |     | NULL    |                |
| updated_at                 | timestamp        | YES  |     | NULL    |                |
+----------------------------+------------------+------+-----+---------+----------------+

這是第一個查詢的查詢結果：

+----+---------+--------------+---------------------+---------------------+
| id | user_id | is_completed | created_at          | updated_at          |
+----+---------+--------------+---------------------+---------------------+
|  6 |       1 |            1 | 2017-05-20 18:04:08 | 2017-05-24 17:35:37 |
| 11 |       1 |            1 | 2017-05-24 22:54:56 | 2017-05-24 22:55:28 |
| 12 |       1 |            1 | 2017-05-24 23:12:09 | 2017-05-24 23:12:38 |
+----+---------+--------------+---------------------+---------------------+

以及來自order_payments表的結果：

+----+----------+------------+
| id | order_id | partner_id |
+----+----------+------------+
|  7 |        6 |          2 |
| 12 |       11 |          2 |
| 14 |       12 |          2 |
+----+----------+------------+

並且，為了使事情更加有趣，此查詢確實起作用並成功返回結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '2') and `id` = '6';

因此，訂單ID 6有效，11不起作用，12不起作用。

這是order_payments表的全部內容：

+----+----------+------------+
| id | order_id | partner_id |
+----+----------+------------+
|  1 |        1 |          1 |
|  2 |        2 |          1 |
|  3 |        3 |          1 |
|  4 |        4 |          1 |
|  5 |        5 |          1 |
|  6 |        6 |          1 |
|  7 |        6 |          2 |
|  8 |        7 |          1 |
|  9 |        8 |          1 |
| 10 |        9 |          1 |
| 11 |       10 |          1 |
| 12 |       11 |          2 |
| 13 |       11 |          1 |
| 14 |       12 |          2 |
| 15 |       12 |          1 |
+----+----------+------------+

我唯一能想到的可能是一個mysql bug，它以某種方式與order_payments id 6匹配，該訂單ID與6的訂單ID匹配，因為這是order_payments id和orders ID匹配的唯一情況。 這是一個mysql bug還是我需要使用不同的別名？

並且，為了使事情更超級傻瓜，以下是一些更時髦的查詢結果：

1個結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '2') and `id` = '6';

0結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '2') and `id` = '11';

0結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '2') and `id` = '12';

現在，相同的查詢但帶有partner_id的查詢從2更改為1：

1個結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '1') and `id` = '6';

1個結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '1') and `id` = '11';

1個結果：

select * from `orders` where `orders`.`is_completed` = '1' and exists (select * from `order_payments` where `orders`.`id` = `order_payments`.`order_id` and `order_payments`.`partner_id` = '1') and `id` = '12';

這些應該返回有效的結果，因為這三個訂單中的所有兩個都向兩個伙伴都付款了。 時髦的東西！

Answer 1

解決問題之后，通過重新編寫子查詢，在子查詢的選擇中將星號（*）替換為“ id”，從而實現了mysql為查詢提供正確結果的解決方案。

select * 
from `orders` 
where `is_completed` = '1' 
and exists (
        select `id` 
        from `order_payments` 
        where `orders`.`id` = `order_payments`.`order_id` 
        and `partner_id` = '2'
) 
and `orders`.`id` = '12'

即使這解決了問題，我仍然認為這是mysql中的錯誤，阻止了原始查詢提供正確的結果。