如何根据 SQL/php 中另一列的多个值从一列中选择单个值

Question

数据库不是我的。结构是

fid| uid | value
 3 |  3  | spain
 3 |  5  | France
 2 |  3  | 45
 6 |  3  | male
 6 |  5  | female
 2 |  5  | 32

字段 ID 是另一个表中的主键，我想忘记。 我试图找到所有具有值“西班牙”、“男性”的 uid

我有以下工作。

SELECT uid 
FROM DATABASE
WHERE value IN ('spain', 'male') GROUP BY uid HAVING COUNT(*) >= 2

问题如下..我如何选择男性，西班牙，价值在 20-30 范围内 （这是在西班牙出生、年龄在 20-30 岁之间的男性吗？

非常感谢！！

Answer 1

你的桌子真的很乱。你把不相关的东西的价值混在一起了。

尝试：

SELECT uid
FROM DATABASE a JOIN
  DATABASE b USING (uid)
WHERE a.value IN ('spain', 'male')
  AND b.value >= 20
  AND b.value <= 30
GROUP BY uid
HAVING COUNT(*) >= 2

请注意，我正在比较字符串和整数。您需要进行测试以了解其效果如何。

另外，我只是把西班牙和男性放在一起，但也许它们真的无关？

fid 决定值的类型吗？

尝试：

SELECT uid
FROM DATABASE country
JOIN DATABASE gender USING (uid)
JOIN DATABASE age USING (uid)
WHERE
  country.fid = 3 AND
  gender.fid = 6 AND
  age.fid = 2 AND
  county.value = 'spain' AND
  gender.value = 'male' AND
  age.value >= 20
  age.value <= 30
GROUP BY uid
HAVING COUNT(*) >= 2

这段代码应该工作得更可靠。

Answer 2

使用自连接：

SELECT tland.uid
FROM `table` AS tland
INNER JOIN `table` AS tgender ON tland.uid = tgender.uid
INNER JOIN `table` AS tage ON tland.uid = tage.uid
WHERE tland.value = 'spain'
  AND tgender.value = 'male'
  AND 20 <= tage.value AND tage.value <= 30

Answer 3

这会起作用。

SELECT d1.uid FROM demo d1
INNER JOIN demo d2 ON d2.uid = d1.uid AND d2.fid = 6 AND d2.value = "Male"
INNER JOIN demo d3 ON d3.uid = d1.uid AND d3.fid = 3 AND d3.value = "Spain"
WHERE (d1.fid = 2 AND (d1.value BETWEEN 20 AND 30))

Answer 4

保持语法尽可能接近原来的语法，这可能会起作用：

SELECT uid
FROM database
WHERE value in ('spain', 'male') or convert(value, , SIGNED INTEGER) between 30 and 50
GROUP BY uid HAVING COUNT(*) >= 3

Answer 5

由于您的字段是这样映射的：

fid 3 = country
fid 2 = age
fid 6 = sex

您可以将该表转换为更逻辑的视图（在 UID 上索引）。

(SELECT c.uid, c.country, a.age, s.sex FROM
   (SELECT uid, value AS country FROM maintable WHERE fid = 3) c INNER JOIN
   (SELECT uid, value AS age FROM maintable WHERE fid = 2) a ON c.uid = a.uid INNER JOIN
   (SELECT uid, value AS sex FROM maintable WHERE fid = 6) s ON c.uid = s.uid
) as X

一旦它位于视图内，您就可以将数据视为常规表。我假设您可能有超过 3 种类型的数据，因为在您的示例中 fid 最多为 6。

假设您已将其放入视图（称为 X）中，您可以执行以下操作：

SELECT * FROM X
WHERE age BETWEEN 20 AND 30
   AND country = 'spain'
   AND sex = 'male'