MySQL

Question

在 MySQL 数据库中，我有 3 个 table：customers、projects 和 tasks。对于我正在实施的搜索，我想搜索所有 3 个 table 和 select 找到的匹配项。问题是我想同样限制 MySQL.

返回的结果数量

这是我当前查询的示例：

SELECT id, title, type
FROM (
    (
    SELECT id, title, 'customer' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM customers
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
    UNION DISTINCT
    (
    SELECT id, title, 'project' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM projects
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
    UNION DISTINCT
    (
    SELECT id, title, 'task' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM tasks
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
) res
LIMIT 6;

在此示例中，我想将结果限制为 6。

我希望此示例的最终结果如下：

1) 如果所有 table 至少有 2 个结果，则每个显示 2 个结果。

id     title                      type
20     'First test customer'      'customer'
22     'Test customer 2'          'customer
48     'A project for testing'    'project'
17     'Test Project'             'project'
1      'Task test'                'task'
2      'Second test'              'task'

2) 如果一个 table 没有任何结果，则为其他 2 个 table 中的每一个显示 3 个结果。（如果只有一个 table 有结果，则显示 table 的 6 个结果。）

id     title                      type
20     'First test customer'      'customer'
22     'Test customer 2'          'customer
56     'Customer test 56'         'customer'
1      'Task test'                'task'
2      'Second test'              'task'
3      'Test task'                'task'

3) 如果 table 中的 2 个有超过 2 个结果，而第三个 table 只有 1 个结果，则显示 table 之一的 3 个结果足够的结果，两个中的另一个有 2 个结果，table 有 1 个结果，只有 1 个结果。

id     title                      type
20     'First test customer'      'customer'
48     'A project for testing'    'project'
17     'Test Project'             'project'
34     'Testing project'          'project'
1      'Task test'                'task'
2      'Second test'              'task'

谁能帮我解决这个问题？

提前致谢！

Answer 1

您可以为每个 SELECT 使用一个行号，然后使用该计算字段对 UNION 进行排序以平衡从单个查询获得的结果（我没有测试过此代码，请以此为起点):

SET @rank1=0;
SET @rank2=0;
SET @rank3=0;
SELECT id, title, type, rank
FROM (
    (

    SELECT @rank1:=@rank1+1 AS rank, id, title, 'customer' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM customers
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
    UNION DISTINCT
    (

    SELECT @rank2:=@rank2+1 AS rank, id, title, 'project' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM projects
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
    UNION DISTINCT
    (

    SELECT @rank3:=@rank3+1 AS rank, id, title, 'task' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM tasks
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
) res
ORDER BY rank 
LIMIT 6;

Answer 2

要在不声明@rankX 变量的情况下使@kiks73 在一个查询中产生结果，您可以在 from 子句中添加 (SELECT @rankX:=0) AS t。像这样：

SELECT id, title, type, rank
FROM (
    (

    SELECT @rank1:=@rank1+1 AS rank, id, title, 'customer' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM customers,(SELECT @rank1:=0) AS t
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
    UNION DISTINCT
    (

    SELECT @rank2:=@rank2+1 AS rank, id, title, 'project' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM projects,(SELECT @rank2:=0) AS t
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
    UNION DISTINCT
    (

    SELECT @rank3:=@rank3+1 AS rank, id, title, 'task' AS type, MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE) AS score
    FROM tasks,(SELECT @rank3:=0) AS t
    WHERE MATCH (title) AGAINST ('+test* ' IN BOOLEAN MODE)
    )
) res
ORDER BY rank
LIMIT 6;

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