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Trying to get a tricky mysql select statement working - and I need some help to cool off my burning noggin...and I have a feeling one of you MYSQL heroes out there will look at this and reel it off.

Goal: List a given user's songs NOT in a given category.

Three tables: : table 1: song, many fields with assigned UserID, and unique SongID

table 2: category, 3+ fields, with assigned UserID and unique CatID

table 3: linker, one-to-many for listing songs in one or more categories. 3 fields, unique id (index), SongID, CatID

The following gets me close - but does not list a user's songs that aren't assigned to any other category at all OR are a already assigned to a another category (I think thanks to !=). 

SELECT DISTINCT song.SongName, song.UserID, song.SongID  FROM song  
JOIN linker ON song.SongID = linker.SongID 
WHERE  linker.CatID != 155 AND song.UserID =2

Then, I tried this, 

SELECT DISTINCT song.SongName, song.UserID, song.SongID FROM song 
LEFT JOIN linker ON song.SongID = linker.SongID 
WHERE (linker.SongID IS NULL OR linker.CatID != 155) AND song.UserID =2

Closer but not working (still thanks to != including songs already assigned).

I was thinking I can get away without invoking table 2, since it merely adds and defines categories for a given user. Alternatively, I'm thinking of getting all the user's songs, and then unsetting array values with a given CatID - but this doesn't seem like it should be necessary? I feel like I'm missing something simple? Table structure is not sacred at this point if it absolutely needs to change. Thanks to any who share their thoughts.

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2 回答 2

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Try this (I am used to MSSQL so if my syntax is off, appologies in advance):

SELECT s.SongName, s.UserID, s.SongID
FROM song s
LEFT JOIN linker l on s.SongID = l.SongID AND l.CatID = 155
WHERE s.UserID = 2
AND l.ID is null
于 2013-09-23T17:34:34.247 回答
0 
SELECT DISTINCT song.SongName, song.UserID, song.SongID 
  FROM song 
  LEFT JOIN linker 
    ON song.SongID = linker.SongID 
   and linker.CatID != 155 
   AND song.UserID = 2
于 2013-09-23T17:20:32.217 回答