Question

我有两张表有一对多的关系。我通过id列加入表。我的问题是我需要计算第二个（tablekey_id）表中所有匹配的条目，但我需要来自标有布尔is_basedomain的行的信息。作为注释，每行只有一行is_basedomain = 1，具有相同的tablekey_id。

表：tablekey

id  linkdata_id  timestamp
22  9495028175   2013-03-10 01:13:46
23  8392740179   2013-03-10 21:23:25

表：searching_domains。

注意：tablekey_id是tablekey表中id的外键。

id  tablekey_id  domain        is_basedomain
1   22           somesite.com  1
2   22           yahoo.com     0
3   23           red.com       1
4   23           blue.com      0
5   23           green.com     0

继承我正在使用的查询。我试图使用子查询，但我似乎只能选择当前tablekey_id的计数，所以这不起作用。

SELECT `tablekey_id`, `linkdata_id`, `timestamp`, `domain`, `is_basedomain`,
   (SELECT COUNT(1) AS other FROM `searched_domains` AS dd 
    ON dd.tablekey_id = d.tablekey_id GROUP BY `tablekey_id`) AS count
FROM `tablekey` AS k 
JOIN `searched_domains` AS d
ON k.id = d.tablekey_id 
WHERE `is_basedomain` = 1 GROUP BY `tablekey_id`

我想回来的结果是：

tablekey_id  linkdata_id  timestamp            domain        is_basedomain    count
22           9495028175   2013-03-10 01:13:46  somesite.com  1                2
23           8392740179   2013-03-10 21:23:25  red.com       1                3

任何人都可以帮助我将其纳入一个查询吗？

Answer 1

您可以将search_domains中包含is_basedomain = 1的行视为查询中的单独表格，并将其与另一个search_domains实例连接（以获取计数）：

SELECT 
    d.tablekey_id, 
    k.linkdata_id, 
    k.timestamp,  
    d.domain, 
    d.is_basedomain, 
    COUNT(*) as 'count'
FROM 
    tablekey AS k
      join searched_domains AS d on d.tablekey_id=k.id
      join searched_domains AS d2 on d2.tablekey_id=d.tablekey_id
WHERE 
    d.is_basedomain = 1 
GROUP BY 
    d.tablekey_id, 
    k.linkdata_id, 
    k.timestamp,  
    d.domain, 
    d.is_basedomain

Answer 2

使用ON代替WHERE

时出错

试试这个

    SELECT `tablekey_id`, `linkdata_id`, `timestamp`, `domain`, `is_basedomain`,
    (SELECT COUNT(1) AS other FROM `searched_domains` AS dd 
    where dd.tablekey_id = d.tablekey_id GROUP BY `tablekey_id`) AS count

    FROM `tablekey` AS k 
    JOIN `searched_domains` AS d
    ON k.id = d.tablekey_id 
    WHERE `is_basedomain` = 1 GROUP BY `tablekey_id`

DEMO HERE

Answer 3

没有理由使用子查询，或者您的意见是什么？

SELECT 
    `tablekey_id`, 
    `linkdata_id`, 
    `timestamp`,  
    `domain`, 
    `is_basedomain`, 
    COUNT(*) as count
FROM 
    `tablekey` AS k , 
    `searched_domains` AS d
WHERE 
    k.id = d.tablekey_id AND 
    `is_basedomain` = 1 
GROUP BY 
    `tablekey_id`, 
    `linkdata_id`, 
    `timestamp`, 
    `domain`,
    `is_basedomain`

如果您只想使用最新时间戳MAX(timestamp) as timestamp，请将其从group by中删除。

选择具有连接表的行数

3 个答案: