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一對多與三個表聯接

[英]One to many join with three tables

我有一個表:網站,廣告活動,並out了運動追蹤系統我建立。 單擊鏈接后,out表將在網站ID和廣告系列ID匹配的位置更新匹配。

在out表中,有一個campaign_id和site_id,它們分別對應於sites和campaigns表。 更復雜的是,每個站點可以有4個廣告系列(campaign_a,campaign_b,campaign_id_reviews,campaign_id_reviews_phone)。 我想加入三個表,對於每個站點,我希望將其放在一行上:

site.site_name, site.campaign_id_a, campaigns.campaign_name, out.hits, 
site.campaign_id_b, campaigns.campaign_name, out.hits, 
site.campaign_id_reviews, campaigns.campaign_name, out.hits, 
site.campaign_id_reviews_phone, campaigns.campaign_name, out.hits

這是我的嘗試,它沒有帶回所有site_id / campaign_id組合,它僅為每個site_id帶回一條記錄,而不是所有site_id / campaign_id組合

SELECT s.*, c.*, o.* FROM sites s
INNER JOIN campaigns c ON s.campaign_id_a=c.campaign_id
INNER JOIN campaigns ON s.campaign_id_b=campaigns.campaign_id
INNER JOIN `out` o ON s.campaign_id_a=o.campaign_id AND s.site_id=o.site_id
WHERE s.site_id NOT IN(100,101)
ORDER BY o.site_id ASC

我的具有3條記錄轉儲的Create表:

CREATE TABLE IF NOT EXISTS `sites` (
  `site_id` mediumint(4) NOT NULL AUTO_INCREMENT,
  `site_name` varchar(70) NOT NULL,
  `campaign_id_a` tinyint(4) NOT NULL,
  `campaign_id_b` tinyint(4) NOT NULL,
  `a_display_name` varchar(50) NOT NULL,
  `b_display_name` varchar(50) NOT NULL,
  `campaign_id_reviews` tinyint(4) NOT NULL,
  `campaign_id_reviews_phone` tinyint(3) NOT NULL DEFAULT '4',
  PRIMARY KEY (`site_id`),
  UNIQUE KEY `site_id` (`site_id`),
  UNIQUE KEY `site_name` (`site_name`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=102 ;

INSERT INTO `sites` (`site_id`, `site_name`, `campaign_id_a`, `campaign_id_b`, `a_display_name`, `b_display_name`, `campaign_id_reviews`, `campaign_id_reviews_phone`) VALUES
(1, 'example.com', 1, 8, 'hard456', 'easy123', 3, 4),
(2, 'example.org', 1, 8, 'hard456', 'easy123', 3, 4),
(3, 'example.net', 8, 8, 'easy123', 'easy123', 3, 4);



CREATE TABLE IF NOT EXISTS `out` (
  `out_id` mediumint(7) NOT NULL AUTO_INCREMENT,
  `site_id` tinyint(4) NOT NULL,
  `campaign_id` tinyint(4) NOT NULL DEFAULT '0',
  `hits` int(11) NOT NULL,
  `date_last_hit` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  PRIMARY KEY (`out_id`),
  UNIQUE KEY `site2campaign` (`site_id`,`campaign_id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=101 ;

INSERT INTO `out` (`out_id`, `site_id`, `campaign_id`, `hits`, `date_last_hit`) VALUES
(19, 60, 3, 418, '2015-11-16 22:52:33'),
(10, 2, 1, 1135, '2015-11-15 04:51:32'),
(20, 60, 1, 1710, '2015-11-14 13:52:20');




CREATE TABLE IF NOT EXISTS `campaigns` (
  `campaign_id` tinyint(4) NOT NULL AUTO_INCREMENT,
  `campaign_name` varchar(60) NOT NULL,
  `network` varchar(60) NOT NULL,
  `url` varchar(400) NOT NULL,
  PRIMARY KEY (`campaign_id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=10 ;

INSERT INTO `campaigns` (`campaign_id`, `campaign_name`, `network`, `url`) VALUES
(1, 'Hard456', 'Hard Network', 'exampleURL'),
(3, 'medium678', 'Medium Network', 'examplewithURL'),
(8, 'easy123', 'Easy Network', 'exampleURLLoaction');
(4, 'none23', 'None Network', 'urlExample');

問題是您要覆蓋列名。 盡管我還沒有測試過,但這應該可行。 您可以使用類似的模式來獲取每個廣告系列ID的out.hits。

SELECT
    r3.*, c4.campaign_name c_name_rev_phone
FROM
    (SELECT
            r2.*, c3.campaign_name c_name_rev
    FROM
        (SELECT 
            r1.*, c2.campaign_name c_name_b
        FROM
            (SELECT 
                s1.site_id s_id, 
                s1.site_name sname, 
                s1.campaign_id_a c_id_a, 
                s1.campaign_id_b c_id_b,
                s1.campaign_id_reviews c_id_rev,
                s1.campaign_id_reviews_phone c_id_rev_phone,
                c1.campaign_name c_name_a 
            FROM sites s1 JOIN campaigns c1 ON s1.campaign_id_a = c1.campaign_id ) r1
        JOIN campaigns c2 ON c2.campaign_id=r1.c_id_b ) r2
    JOIN campaigns c3 ON c3.campaign_id=r2.c_id_rev ) r3
JOIN campaigns c4 on c4.campaign_id=r3.c_id_rev_phone;

編輯1-

根據問題中提供的樣本數據,以下查詢的結果將是

SELECT s.*, c.* FROM sites s
INNER JOIN campaigns c ON s.campaign_id_a=c.campaign_id
INNER JOIN campaigns ON s.campaign_id_b=campaigns.campaign_id
WHERE s.site_id NOT IN(100,101);

結果:

site_id |site_name      |campaign_id_a  |campaign_id_b  |a_display_name |b_display_name |campaign_id_reviews    |campaign_id_reviews_phone  |campaign_id    |campaign_name  |network        |url
--------|---------------|---------------|---------------|---------------|---------------|-----------------------|---------------------------|---------------|---------------|---------------|------------------
1       |example.com    |1              |8              |hard456        |easy123        |3                      |4                          |1              |Hard456        |Hard Network   |exampleURL
2       |example.org    |1              |8              |hard456        |easy123        |3                      |4                          |1              |Hard456        |Hard Network   |exampleURL
3       |example.net    |8              |8              |easy123        |easy123        |3                      |4                          |8              |easy123        |Easy Network   |exampleURLLoaction

答案查詢結果:

s_id    |sname          |c_id_a |c_id_b |c_id_rev   |c_id_rev_phone |c_name_a   |c_name_b   |c_name_rev |c_name_rev_phone
------------------------|-------|-------|-----------|---------------|-----------|-----------|-----------|----------------
1       |example.com    |1      |8      |3          |4              |Hard456    |easy123    |medium678  |none23
2       |example.org    |1      |8      |3          |4              |Hard456    |easy123    |medium678  |none23
3       |example.net    |8      |8      |3          |4              |easy123    |easy123    |medium678  |none23

請注意,如何分別獲取campaign_id_acampaign_id_bcampaign_id_reviews等的廣告系列名稱。

這是可以為您提供完整答案的查詢:

SELECT
r3.*, 
c4.campaign_name c_name_rev_phone,
o4.hits hits_rev_phone
FROM
    (SELECT
            r2.*, 
            c3.campaign_name c_name_rev,
            o3.hits hits_rev
    FROM
        (SELECT 
            r1.*, 
            c2.campaign_name c_name_b, 
            o2.hits hits_b
        FROM
            (SELECT 
                s1.site_id s_id, 
                s1.site_name sname, 
                s1.campaign_id_a c_id_a, 
                s1.campaign_id_b c_id_b,
                s1.campaign_id_reviews c_id_rev,
                s1.campaign_id_reviews_phone c_id_rev_phone,
                c1.campaign_name c_name_a,
                o1.hits hits_a
            FROM sites s1 JOIN campaigns c1 ON s1.campaign_id_a = c1.campaign_id JOIN `out` o1 ON (c1.campaign_id=o1.campaign_id AND o1.site_id=s1.site_id)) r1
        JOIN campaigns c2 ON c2.campaign_id=r1.c_id_b JOIN `out` o2 ON (c2.campaign_id=o2.campaign_id AND o2.site_id=r1.s_id)) r2
    JOIN campaigns c3 ON c3.campaign_id=r2.c_id_rev JOIN `out` o3 ON (c3.campaign_id=o3.campaign_id AND o3.site_id=r2.s_id)) r3
JOIN campaigns c4 on c4.campaign_id=r3.c_id_rev_phone JOIN `out` o4 ON (c4.campaign_id=o4.campaign_id AND o4.site_id=r3.s_id);

這將為您提供的示例數據集提供空白結果,因為表out中不存在很多campaign_idsite_id組合。 由於我們正在進行內部INNER您將失去這些信息。 如果您希望在site_idcampaign_id組合不存在時將hits報告為0 ,則需要使用LEFT JOIN

如果這不是您想要的,請隨時還原。

您似乎兩次參加了競選活動,但只返回了與compaign_id_a相匹配的任何內容。 結果中會忽略Campaign_id_b,而其他兩個廣告系列ID都不會處理。

將其拆分以獲取每個廣告系列ID,並將結果合並在一起:-

(SELECT s.site_id,
        s.site_name,
        s.campaign_id_a,
        s.campaign_id_b,
        s.a_display_name,
        s.b_display_name,
        s.campaign_id_reviews,
        s.campaign_id_reviews_phone,
        o.out_id,
        o.hits,
        o.date_last_hit, 
        c.campaign_id,
        c.campaign_name,
        c.network,
        c.url 
FROM sites s
INNER JOIN campaigns c ON s.campaign_id_a = c.campaign_id
INNER JOIN `out` o ON c.campaign_id = o.campaign_id AND s.site_id = o.site_id
WHERE s.site_id NOT IN (100,101))
UNION
(SELECT s.site_id,
        s.site_name,
        s.campaign_id_a,
        s.campaign_id_b,
        s.a_display_name,
        s.b_display_name,
        s.campaign_id_reviews,
        s.campaign_id_reviews_phone,
        o.out_id,
        o.hits,
        o.date_last_hit, 
        c.campaign_id,
        c.campaign_name,
        c.network,
        c.url 
FROM sites s
INNER JOIN campaigns c ON s.campaign_id_b = c.campaign_id
INNER JOIN `out` o ON c.campaign_id = o.campaign_id AND s.site_id = o.site_id
WHERE s.site_id NOT IN (100,101))
UNION
(SELECT s.site_id,
        s.site_name,
        s.campaign_id_a,
        s.campaign_id_b,
        s.a_display_name,
        s.b_display_name,
        s.campaign_id_reviews,
        s.campaign_id_reviews_phone,
        o.out_id,
        o.hits,
        o.date_last_hit, 
        c.campaign_id,
        c.campaign_name,
        c.network,
        c.url 
FROM sites s
INNER JOIN campaigns c ON s.campaign_id_reviews = c.campaign_id
INNER JOIN `out` o ON c.campaign_id = o.campaign_id AND s.site_id = o.site_id
WHERE s.site_id NOT IN (100,101))
UNION
(SELECT s.site_id,
        s.site_name,
        s.campaign_id_a,
        s.campaign_id_b,
        s.a_display_name,
        s.b_display_name,
        s.campaign_id_reviews,
        s.campaign_id_reviews_phone,
        o.out_id,
        o.hits,
        o.date_last_hit, 
        c.campaign_id,
        c.campaign_name,
        c.network,
        c.url 
FROM sites s
INNER JOIN campaigns c ON s.campaign_id_reviews_phone = c.campaign_id
INNER JOIN `out` o ON c.campaign_id = o.campaign_id AND s.site_id = o.site_id
WHERE s.site_id NOT IN (100,101))
ORDER BY site_id ASC

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