MySQL 5.5.43
我正在研究7200种大麻菌株的数据库,需要显示菌株列表以及其育种者声称的最受欢迎的物种。
该主题非常令人困惑,因此以下一些事实将帮助您了解我的困惑所在:
示例应变:
对于一种非常受欢迎的菌株White Widow,这是我产生的结果集。它有29个不同的育种者,每个育种者声称拥有不同的物种。从结果中可以看到,此菌株最流行的物种是In /苜蓿(同等杂种)。
SELECT
s.id,
b.id AS breederID,
b.breederName AS breederName,
GROUP_CONCAT(DISTINCT sp.species ORDER BY sp.species ASC SEPARATOR '/') AS species
FROM strains AS s
LEFT JOIN strainBreedersDir AS sbd ON s.id = sbd.strainID
LEFT JOIN breeders AS b ON sbd.breederID = b.id
LEFT JOIN strainBreederSpeciesDir AS sbsd ON s.id = sbsd.strainID AND sbd.breederID = sbsd.breederID
LEFT JOIN species AS sp ON sbsd.speciesID = sp.id
WHERE s.id = 6782
GROUP BY s.id, sbd.breederID
我想要的结果
我想显示一个品系名称列表,然后在每个名称旁边显示一个育种者列表,以及所有育种者最受欢迎的/平均声称的物种。因此,正如我之前向您展示的,育种者为此菌株记录的最受欢迎的物种是印度In /苜蓿,并希望这样显示:
strainID | strainName | breeders | averageSpecies
--------------------------------------------------------------------------
6782 | White Widow | Green House Seeds, | Indica/Sativa
| | Barney's Farm
我尝试过的是:
我没有在每个物种旁边显示最受欢迎的物种,而是在每个物种旁边显示了记录物种的第一个实例。我以为这样就足够了,但是一个物种的第一个实例可能是空的,因为目前大约有100个未知物种的菌株。因此,我不希望某个物种的第一个实例是“未知的”,因为实际上该菌株的其他育种者知道其中的物种。因此,我认为最好是确定记录最多的物种并显示出来。到目前为止,这是我必须去的地方:
SELECT
s.id,
s.strainName,
GROUP_CONCAT(DISTINCT b.breederName ORDER BY b.breederName ASC separator ', ') AS breeders,
COALESCE(NULLIF(ps.primarySpecies,''),'Unknown') AS primarySpecies
FROM strains AS s
LEFT JOIN strainBreedersDir AS sbd ON s.id = sbd.strainID
LEFT JOIN breeders AS b ON sbd.breederID = b.id
LEFT OUTER JOIN (
SELECT
sbd.breederID AS breederID,
GROUP_CONCAT(DISTINCT sp.species ORDER BY sp.species ASC SEPARATOR '/') AS primarySpecies
FROM strains AS s
LEFT JOIN strainBreedersDir AS sbd ON s.id = sbd.strainID
LEFT JOIN strainBreederSpeciesDir AS sbsd ON s.id = sbsd.strainID AND sbd.breederID = sbsd.breederID
LEFT JOIN species AS sp ON sbsd.speciesID = sp.id
GROUP BY s.id, sbd.breederID
) AS ps ON sbd.breederID = ps.breederID
WHERE s.id = 6782
GROUP BY s.id
结果
id | strainName | breeders | species
----------------------------------------------------------
6782 | White Widow | Green House Seeds, | Indica/Sativa
| | Barney's Farm |
但是我无法弄清楚如何修改OUTER JOIN
来显示最受欢迎的物种,而不仅仅是第一行。我尝试了许多不同的外部联接查询变体,但都失败了很多,而对所尝试的方法却一无所知。
如何显示最受欢迎的物种?
数据库结构:
strains
id (PK AUTO) | strainName (UNIQUE)
---------------------------------------------
6782 | White Widow
-
strainBreedersDir
strainID (FK UNIQUE) | breederID (UNIQUE)
---------------------------------------------
6782 | 16
6782 | 23
-
breeders
id (PK AUTO) | breederName (UNIQUE)
---------------------------------------------
16 | Green House Seeds
23 | Barney's Farm
-
strainBreederSpeciesDir
strainID (FK UNIQUE) | breederID (INT UNIQUE) | speciesID (INT UNIQUE)
----------------------------------------------------------------------
6782 | 16 | 1
6782 | 16 | 2
6782 | 23 | 5
-
species
id (PK AUTO) | species (UNIQUE)
-------------------------------------
1 | Indica
2 | Sativa
3 | Ruderalis
4 | Mostly Indica
5 | Mostly Sativa
6 | Mostly Ruderalis
这是一个SQLFIDDLE-由Juan Carlos Oropeza提供。
我不了解您要从工作查询中汇总的知识。
ID可能以不同的方式执行此操作,但是由于我没有更改您的工作查询,因此可能会为您提供所需的信息。子查询的出现GROUP_CONCAT
使事情变得有些棘手,因为我们正在指望那个领域,而我不能直接把它放在那儿(除非有人可以向我展示更好的方法),然后我从中选择aMAX
或AVG
。您可以切换MAX
为AVG
。
SELECT MAX(aggregated.theCount),
aggregated.id,
aggregated.breederID,
aggregated.breeders as mostPopularBreeders,
aggregated.species as mostPopularSpecies,
AllStrainBreeders.allBreeders as strainBreeders
FROM(
SELECT
speciesWithBreeder.id,
speciesWithBreeder.breederID,
speciesWithBreeder.breederName,
GROUP_CONCAT(DISTINCT speciesWithBreeder.breederName ORDER BY speciesWithBreeder.breederName ASC separator ', ') AS breeders,
speciesWithBreeder.species,
COUNT(*) as theCount
FROM(
SELECT
s.id,
b.id AS breederID,
b.breederName AS breederName,
GROUP_CONCAT(DISTINCT sp.species ORDER BY sp.species ASC SEPARATOR '/') AS species
FROM strains AS s
LEFT JOIN strainBreedersDir AS sbd ON s.id = sbd.strainID
LEFT JOIN breeders AS b ON sbd.breederID = b.id
LEFT JOIN strainBreederSpeciesDir AS sbsd ON s.id = sbsd.strainID AND sbd.breederID = sbsd.breederID
INNER JOIN species AS sp ON sbsd.speciesID = sp.id
WHERE s.id = 6782
GROUP BY s.id, sbd.breederID)
AS speciesWithBreeder
GROUP BY speciesWithBreeder.species
ORDER BY COUNT(*) DESC
) as aggregated
LEFT JOIN(
SELECT
sbd.strainID,
GROUP_CONCAT(DISTINCT b.breederName ORDER BY b.breederName ASC SEPARATOR ',') AS allBreeders
FROM breeders b
LEFT JOIN strainBreedersDir sbd ON sbd.breederID = b.id AND sbd.strainID = 6782
GROUP BY sbd.strainID
) as AllStrainBreeders
ON aggregated.id = AllStrainBreeders.strainID
GROUP BY aggregated.id
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