MySql查询获取每个国家为每个目的花费的天数？ （获得第一个表中第二个表中所有记录的计数）

Question

我有三个表tl_log，tl_geo_countries，tl_purpose。我试图在表'tl_purpose'中为每个目的获得表'tl_log'中每个国家花费的天数。 我下面试过mysql查询

SELECT t.country_id AS countryID,t.reason_id AS reasonID,count(t.reason_id) AS 
days,c.name AS country, p.purpose AS purpose 
FROM `tl_log` AS t 
LEFT JOIN tl_geo_countries AS c ON t.country_id=c.id 
LEFT JOIN tl_purpose AS p ON t.reason_id=p.id 
GROUP BY t.reason_id,t.country_id ORDER BY days DESC 

但是登陆了。

我不能够得到计数宗旨，为在“tl_log”中不存在表“tl_log”每一个国家。任何帮助是极大的赞赏。另外，如果问题难以理解，请让我知道。

预期输出：

下面是这三个表的结构

tl_log

个tl_geo_countries

tl_purpose

Answer 1

被替换如果你想国家和宗旨的所有可能的组合，即使是那些不出现在日志表上（这些将显示为机智）哈0计数），你可以先做两个表的笛卡尔积（一CROSS加入），然后LEFT加入到日志表：

SELECT 
    c.id AS countryID, 
    p.id AS reasonID, 
    COUNT(t.reason_id) AS days, 
    c.name AS country, 
    p.purpose AS purpose 
FROM 
    tl_geo_countries AS c 
    CROSS JOIN 
    tl_purpose AS p 
    LEFT JOIN 
    tl_log AS t 
     ON t.country_id = c.id 
     AND t.reason_id = p.id 
GROUP BY 
    p.id, 
    c.id 
ORDER BY 
    days DESC ; 

如果你想记录只存在于国家在日志表（但仍然是所有可能的原因/目的），需要稍作修改：

SELECT 
    c.id AS countryID, 
    p.id AS reasonID, 
    COUNT(t.reason_id) AS days, 
    c.name AS country, 
    p.purpose AS purpose 
FROM 
    (SELECT DISTINCT 
      country_id 
     FROM 
      tl_log 
    ) AS dc 
    JOIN 
    tl_geo_countries AS c 
     ON c.id = dc.country_id 
    CROSS JOIN 
    tl_purpose AS p 
    LEFT JOIN 
    tl_log AS t 
     ON t.country_id = c.id 
     AND t.reason_id = p.id 
GROUP BY 
    p.id, 
    c.id 
ORDER BY 
    days DESC ; 

Answer 2

LEFT JOIN应RIGHT JOIN