MySQL, three tables: Select all rows in right table including rows that are not mapped in middle table

Question

My schema are as follows:

Sites S
| S.Id | S.Url |
| 1    | a.com |
| 2    | b.edu |
| 3    | c.org |

SiteFeatures SF
| SF.SiteId | SF.FeatureID |
| 1         | 1            |
| 1         | 2            |
| 1         | 3            |
| 2         | 1            |
| 2         | 2            |
| 2         | 3            |
| 2         | 4            |
| 3         | 2            |
| 3         | 3            |

Features F
| F.Id | F.FeatureName |
| 1    | apple         |
| 2    | banana        |
| 3    | cherry        |
| 4    | diaper        |
| 5    | egg           |
| 6    | fish          |

I want to select all sites, mapped to all features, including features missing from the middle join table. For features that have no entry in the join table, I want to display a "0". For features that exist in the join table, I want a "1".

So the results will look like this:

| SiteId | SiteURL | FeatureName | Enabled |
| 1      | a.com   | apple       | 1       |
| 1      | a.com   | banana      | 1       |
| 1      | a.com   | cherry      | 1       |
| 1      | a.com   | diaper      | 0       |
| 1      | a.com   | egg         | 0       |
| 1      | a.com   | fish        | 0       |
| 2      | b.edu   | apple       | 1       |
| 2      | b.edu   | banana      | 1       |
| 2      | b.edu   | cherry      | 1       |
| 2      | b.edu   | diaper      | 1       |
| 2      | b.edu   | egg         | 0       |
| 2      | b.edu   | fish        | 0       |
| 3      | c.org   | apple       | 0       |
| 3      | c.org   | banana      | 1       |
| 3      | c.org   | cherry      | 0       |
| 3      | c.org   | diaper      | 1       |
| 3      | c.org   | egg         | 0       |
| 3      | c.org   | fish        | 0       |

-- EDIT -- Additional Information.

I originally created a pivot table using this article:

http://dev.mysql.com/tech-resources/articles/wizard/print_version.html

Based on that article, I wrote a SQL statement to dynamically generate a SQL query that generates the pivot table.

That statement looked like this:

SELECT
 CONCAT( ", SUM(IF( F.FeatureName = '" , F.FeatureName , "', 1,0 ))" , " AS `" , F.FeatureName , "` ") AS CutNPaste
FROM
  Features F
WHERE 1
GROUP BY F.FeatureName
ORDER BY F.FeatureName
-- END

The second SQL statement was as follows:

SELECT
, S.Url
/* This section was dynamically generated, and copied into this SELECT statement */
, SUM(IF( F.FeatureName = 'apple', 1,0 )) AS `apple`
, SUM(IF( F.FeatureName = 'banana', 1,0 )) AS `banana`
, SUM(IF( F.FeatureName = 'cherry', 1,0 )) AS `cherry`
, SUM(IF( F.FeatureName = 'diaper', 1,0 )) AS `diaper`
, SUM(IF( F.FeatureName = 'egg', 1,0 )) AS `egg`
, SUM(IF( F.FeatureName = 'fish', 1,0 )) AS `fish`
/* END of dynamic part */
FROM
  Sites S
LEFT OUTER JOIN SiteFeatures SF ON S.Id = SF.SiteId
LEFT OUTER JOIN Features F ON SF.FeatureId = F.Id
WHERE 1
AND SF.FeatureId = F.Id
AND S.Enabled = 1
GROUP BY S.Url
-- END

The results looked like this:

| Url   | apple | banana | cherry | diaper | egg | fish |
| a.com | 1     | 1      | 1      | 0      | 0   | 0    |
| b.edu | 1     | 1      | 1      | 1      | 0   | 0    |
| c.org | 0     | 1      | 0      | 1      | 0   | 0    |

I tried to repurpose the SQL and the concepts in these two statements, but I am at a loss.

What have you tried already - it will help you learn and understand if you have a pop and post your results, we can then help with where you are going wrong, and give you some pointers.. — diagonalbatman, Aug 12 '11 at 13:42
I didn't post what I've tried only because it is rather a mess, and very verbose, and would require me to heavily clean it up. But I'll do that and post back. — JDS, Aug 12 '11 at 14:25

ypercubeᵀᴹ · Answer 1 · 2011-08-12T23:53:21.800

CROSS JOIN can be used here. (The first query was posted first by @nick rulez):

SELECT s.Id   AS SiteId 
     , s.Url  AS SiteURL 
     , f.FeatureName 
     , CASE WHEN sf.SiteID = s.Id 
            THEN 1
            ELSE 0
       END AS Enabled
FROM
    Sites AS s
  CROSS JOIN
    Features AS f
  LEFT JOIN
    SiteFeatures AS sf
      ON  sf.SiteID = s.Id
      AND sf.FeatureID = f.Id

SELECT s.Id   AS SiteId 
     , s.Url  AS SiteURL 
     , f.FeatureName 
     , CASE WHEN EXISTS
                 ( SELECT *
                   FROM SiteFeatures AS sf
                   WHERE sf.SiteID = s.Id
                     AND sf.FeatureID = f.Id
                 ) 
            THEN 1
            ELSE 0
       END AS Enabled
FROM
    Sites AS s
  CROSS JOIN
    Features AS f

SELECT s.Id   AS SiteId 
     , s.Url  AS SiteURL 
     , f.FeatureName 
     , ( SELECT COUNT(*)
         FROM SiteFeatures AS sf
         WHERE sf.SiteID = s.Id
           AND sf.FeatureID = f.Id
       ) AS Enabled
FROM
    Sites AS s
  CROSS JOIN
    Features AS f

+1. Hi ypercube. No problem. Your reputation is growing very fast. Congrats. Nice your second query :) — Nicola Cossu, Aug 12 '11 at 23:55

Sam DeHaan · Answer 2 · 2011-08-12T14:33:20.490

DISCLAIMER: The following solution is written in MS SQL-Server 2008 TSQL.

I don't believe anything I did does not exist in MySQL, but if anything I use is inappropriate in MySQL please let me know so I can correct it. If anyone wishes to edit and correct the syntax to MySQL, you're more than welcome to. I don't know what I need to escape with backticks and whatnot, so I don't want to attempt it and create poor MySQL code out of proper MS Server code.

I also don't pretend to claim that my solution is efficient or elegant, as I am in no way a SQL expert.

The query uses a join to create all possible combinations of Site + Feature. It then checks if that combination exists in the SiteFeature table

--table creation
declare @Sites Table(
    [Id] int not null,
    [Url] varchar(30)
)
INSERT INTO @Sites ([Id],[Url])
     VALUES (1,'a.com'),(2,'b.edu'),(3,'c.org')

declare @Features Table(
    [Id] int not null,
    [Name] varchar(30)
)
INSERT INTO @Features ([Id],[Name])
     VALUES (1,'apple'),(2,'banana'),(3,'cherry'),(4,'diaper'),(5,'egg'),(6,'fish')

declare @SiteFeatures Table(
    [SiteId] int not null,
    [FeatureID] int not null
)
INSERT INTO @SiteFeatures ([SiteId],[FeatureID])
     VALUES (1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(2,4),(3,2),(3,3)

--QUERY HERE
SELECT [SiteId], [SiteURL], [FeatureName], [Enabled] =
    CASE 
        WHEN EXISTS 
            (SELECT * 
             FROM @SiteFeatures SF 
             WHERE SF.SiteId = tmp.SiteId 
                 AND SF.FeatureID = tmp.FeatureId) 
             THEN (1)
             ELSE (0)
        END
    FROM 
    (SELECT S.Id as [SiteId], 
            S.Url as [SiteURL], 
            F.Id as [FeatureId], 
            F.Name as [FeatureName] 
     FROM @Sites S 
    LEFT JOIN @Features F ON (1 = 1)) as tmp

You can use `CROSS JOIN @Features F` instead of `LEFT JOIN @Features F ON (1 = 1)` — ypercubeᵀᴹ, Aug 12 '11 at 23:54
@ypercube thanks. I've got a lot of experience with really basic sql, but haven't had the opportunity to do a lot of... anything else. Good to know, I figured there must be some better way. — Sam DeHaan, Aug 13 '11 at 00:52
And for RDBMS that do not have `CROSS JOIN`, one can simply write: `FROM @Sites S, @Features F` — ypercubeᵀᴹ, Aug 13 '11 at 21:09

Nicola Cossu · Answer 3 · 2011-08-12T22:55:39.237

select s.id,s.url,f.featurename,if(sf.featureid is not null,1,0) as enabled
from features as f
cross join sites as s
left join sitefeatures as sf on sf.siteid = s.id and sf.featureid = f.id
order by s.url,f.featurename;


+----+-------+-------------+---------+
| id | url   | featurename | enabled |
+----+-------+-------------+---------+
|  1 | a.com | apple       |       1 |
|  1 | a.com | banana      |       1 |
|  1 | a.com | cherry      |       1 |
|  1 | a.com | diaper      |       0 |
|  1 | a.com | egg         |       0 |
|  1 | a.com | fish        |       0 |
|  2 | b.edu | apple       |       1 |
|  2 | b.edu | banana      |       1 |
|  2 | b.edu | cherry      |       1 |
|  2 | b.edu | diaper      |       1 |
|  2 | b.edu | egg         |       0 |
|  2 | b.edu | fish        |       0 |
|  3 | c.org | apple       |       0 |
|  3 | c.org | banana      |       1 |
|  3 | c.org | cherry      |       1 |
|  3 | c.org | diaper      |       0 |
|  3 | c.org | egg         |       0 |
|  3 | c.org | fish        |       0 |
+----+-------+-------------+---------+
18 rows in set (0.00 sec)

Ah, I just saw that you have already posted this solution! +1 — ypercubeᵀᴹ, Aug 12 '11 at 23:48
Sweet, thanks. I have a question, though. The MySQL documentation says that "INNER JOIN" and "CROSS JOIN" are the same thing. Can anyone shed some light on that? Ref: http://dev.mysql.com/doc/refman/5.0/en/join.html . "In MySQL, CROSS JOIN is a syntactic equivalent to INNER JOIN (they can replace each other). In standard SQL, they are not equivalent" — JDS, Aug 14 '11 at 13:46

MySQL, three tables: Select all rows in right table including rows that are not mapped in middle table

3 Answers3