Question

我需要帮助：

我有一张名为 phonograms ， phonogram_instruments 和过滤器的表格：

在phonograms表上，我有很多数据可以在我的应用上获取。一个录音制品有许多phonogram_instruments，每个phonogram_instrument都属于一个滤镜。

我需要获取所有有phonogram_instrument.filter_id = 25的phonogram和phonogram_instrument.filter_id = 30

我有一个查询可以正常使用一个值：

SELECT * FROM "phonograms" 
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms" 
   ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
   (phonogram_instruments_phonograms.filter_id = 29)

但是如果我放了另一个值，它就没有了：

 SELECT * FROM "phonograms" 
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms"
   ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
   (phonogram_instruments_phonograms.filter_id = 29)
   AND (phonogram_instruments_phonograms.filter_id = 25)

你可以帮帮我吗？：）

Answer 1

单个filter_id行无法同时具有29和25的值

因此，您应该使用OR而不是AND

SELECT * FROM "phonograms" 
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms" 
ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
 phonogram_instruments_phonograms.filter_id = 29
 OR phonogram_instruments_phonograms.filter_id = 25

如果你的where语句包含更多过滤器，请将filter_id过滤器语句放在两个括号

之间

SELECT * FROM "phonograms" 
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms" 
ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
 (
   phonogram_instruments_phonograms.filter_id = 29
   OR phonogram_instruments_phonograms.filter_id = 25
 )
 AND ... other filters

Answer 2

我的第一个想法是使用这样的INTERSECT复合运算符：

SELECT * FROM "phonograms" 
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms" 
   ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
   phonogram_instruments_phonograms.filter_id = 29
INTERSECT
SELECT * FROM "phonograms" 
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms" 
   ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
   phonogram_instruments_phonograms.filter_id = 25

如果您需要与任意数量的filter_id相交（例如，找到phonogram_instrument.filter_id为29,25和37的所有录音制品），它并不是特别漂亮且不能很好地缩放。在这种情况下，我可能会建议一个有趣的替代方案（顺便提一下，它也适用于两个值）：

SELECT * FROM "phonograms", count(phonograms.id) as 'num_ids'
LEFT OUTER JOIN "phonogram_instruments" "phonogram_instruments_phonograms" 
   ON "phonogram_instruments_phonograms"."phonogram_id" = "phonograms"."id" 
WHERE 
   phonogram_instruments_phonograms.filter_id IN (29, 25)
GROUP BY
   phonograms.id
HAVING
   num_ids = 2

如果您有2个以上的filter_id，请展开WHERE子句中的括号列表，并相应地更改HAVING子句中的数值。

SQL左侧连接的同一列上的2个值的SQL

2 个答案: