Question

假设我有一个像这样的表：

ID|Word |Reference
1 |Dog  |1
1 |Fish |2
1 |Sheep|3
2 |Dog  |4
2 |Fish |5
3 |Sheep|6
4 |Dog  |7

我想选择所有包含单词Dog AND Sheep的ID。所以结果应该是ID：1和2.我尝试使用这个查询：

SELECT ID FROM `Table` WHERE Word='Dog' OR Word='Fish' GROUP BY ID Having Word='Dog AND Word='Fish'

但是，AND子句中的Having使我得到0结果。所以，我做错了什么或者是否有另一种方法来实现我想要的基于MySQL查询的扫描（为了优化速度，因为它必须使用与上例相同的设置搜索许多行）

基本上问题是具有相同ID的多行的AND语句。

更新：
我需要获取找到的ID的引用。例如。返回ID 1和2时，我需要知道ID 1有引用1和2. ID 2有引用3和4.目前，我正在使用此查询：

SELECT ID FROM `Test` WHERE Word in ('Dog', 'Fish') GROUP BY ID HAVING  count(DISTINCT Word) = 2;

由于

Answer 1

以下是两个返回正确记录的解决方案，第一个是ID和参考的单个记录，第二个是每个ID一个记录，单词和参考以逗号分隔列。

设置表并填充行：

DROP TABLE IF EXISTS `list1`;

CREATE table `list1` (
    id int(10),
    Word varchar(10),
    Reference int(10) 
);

INSERT INTO `list1` (`ID`, `Word`, `Reference`) 
VALUES 
(1, 'Dog',1),
(1 ,'Fish',2),
(1 ,'Sheep',3),
(2 ,'Dog',4),
(2 ,'Sheep',5),
(3 ,'Sheep',6),
(4 ,'Dog',7);

为ID和Word的每个组合返回一行

SELECT 
    t.`ID`,
    t.`Word`,
    t.`Reference`
FROM `list1` as t
JOIN (
    SELECT 
        t1.`ID` as `ref_id`
    FROM `list1` AS t1
    WHERE `Word` in ('Sheep','Dog')
    GROUP BY t1.`ID`
    HAVING count(DISTINCT t1.`Word`) = 2
) AS ts
ON t.`ID` = ts.`ref_id`
WHERE t.`Word` in ('Sheep','Dog')
ORDER BY t.`ID`,t.`Word`;

<强>结果

ID  |   Word    |   Reference   
1   |   Dog     |   1
1   |   Sheep   |   3
2   |   Dog     |   4
2   |   Sheep   |   5

每个ID返回一行，在一列中使用逗号分隔的单词列表，在另一列中以逗号分隔的Reference列表。

SELECT 
    t.`ID`,
    GROUP_CONCAT(t.`Word`) AS `Words`,
    GROUP_CONCAT(t.`Reference`) AS `References`
FROM `list1` as t
JOIN (
    SELECT 
        t1.`ID` as `ref_id`
    FROM `list1` AS t1
    WHERE `Word` in ('Sheep','Dog')
    GROUP BY t1.`ID`
    HAVING count(DISTINCT t1.`Word`) = 2
) AS ts
ON t.`ID` = ts.`ref_id`
WHERE t.`Word` in ('Sheep','Dog')
GROUP BY t.`ID`
ORDER BY t.`ID`,t.`Word`;

<强>结果：

ID  |   Words       |   References  
1   |   Dog,Sheep   |   1,3
2   |   Dog,Sheep   |   4,5

Answer 2

您需要自己加入表格。通过这种方式，您可以在狗和羊重叠的情况下获取id相同的位置。试试这个：

declare @t table (id int ,  Word varchar(10) )
insert into @t (ID, Word) values (1, 'Dog'),
(1 ,'Fish'),
(1 ,'Sheep'),
(2 ,'Dog'),
(2 ,'Sheep'),
(3 ,'Sheep'),
(4 ,'Dog')

select t.ID
from @t as t
join @t as t1 on t1.id = t.id
where t.word = 'Dog' and t1.word = 'Sheep'

Answer 3

这是通过将您的桌子加入自身来实现这一目标的一种方式。

SELECT t1.id FROM `Table` t1
INNER JOIN `Table` t2 ON t1.id = t2.id
WHERE t1.word='Dog' AND t2.word='Sheep';

Answer 4

使用下面的查询解决了我的问题的答案：

SELECT ID, GROUP_CONCAT(Reference) as ReferencesGrouped FROM `Test` WHERE Word in ('Dog', 'Fish') GROUP BY ID HAVING  count(DISTINCT Word) = 2;

这会让我回复：

ID|ReferencesGrouped
1 |1,4
2 |4,5

PHP MySQL Group BY有问题

4 个答案: