Question

我创建了一项调查并将其发送出去。该调查询问用户他们的电子邮件，然后要求他们从包含8种不同选项的下拉菜单中选择他们将吃的餐。有些人使用相同的电子邮件多次填写调查，但不同的食物选择。

我有一个如下所示的MySQL数据库：

ID | Email | Type of computer

1 | abc@example.com | Pork
2 | abc@example.com | Chicken
3 | gfh@example.com | Pork
4 | xyz@example.com | Beef

我有超过300,000行的这些数据。我已经制作了一个MySQL查询来组织它，所以它看起来像我上面所示。现在，我想算一下有多少人选择猪肉和鸡肉，有多少人选择猪肉和牛肉等等。有什么帮助吗？

更具体地说，我需要确定数据库中有多少人（有多少个独特的电子邮件）与＃34; Pork＆＃34;和＃34;鸡肉＆＃34;或＆＃34;猪肉＆＃34;和＆＃34;牛肉＆＃34; （等）

Answer 1

要计算选择了Pork和Chicken的人数：

SELECT COUNT(*) AS LikesPorkChicken
FROM MyTable t1
  INNER JOIN Mytable t2 ON t1.Email = t2.Email
WHERE t1.Type ='Pork' 
  AND t2.Type = 'Chicken'

要计算选择了Pork和Beef的人数：

SELECT COUNT(*) AS LikesPorkBeef
FROM MyTable t1
  INNER JOIN Mytable t2 ON t1.Email = t2.Email
WHERE t1.Type ='Pork' 
  AND t2.Type = 'Beef'

如果您想在一个查询中使用它，您可以执行以下操作（SQL Fiddle）：

SELECT SUM(m.PC) AS PorkChicken, SUM(m.PB) AS PorkBeef, SUM(CB) AS ChickenBeef
FROM 
(
  SELECT 
    CASE WHEN t1.Type = 'Pork' AND t2.Type = 'Chicken' THEN 1 ELSE 0 END AS PC,
    CASE WHEN t1.Type = 'Pork' AND t2.Type = 'Beef' THEN 1 ELSE 0 END AS PB,
    CASE WHEN t1.Type = 'Chicken' AND t2.Type = 'Beef' THEN 1 ELSE 0 END AS CB
  FROM MyTable t1
    INNER JOIN Mytable t2 ON t1.Email = t2.Email 
          AND t1.id <> t2.id
) m;

要确保不计算任何重复项，请使用以下（SQL Fiddle）：

SELECT SUM(m.PC) AS PorkChicken, SUM(m.PB) AS PorkBeef, SUM(CB) AS ChickenBeef
FROM 
(
  SELECT 
    CASE WHEN ss.type1 = 'Pork' AND ss.type2 = 'Chicken' THEN 1 ELSE 0 END AS PC,
    CASE WHEN ss.type1 = 'Pork' AND ss.type2 = 'Beef' THEN 1 ELSE 0 END AS PB,
    CASE WHEN ss.type1 = 'Chicken' AND ss.type2 = 'Beef' THEN 1 ELSE 0 END AS CB
  FROM 
  (
    SELECT DISTINCT t1.Email, t1.Type AS type1, t2.Type AS type2
    FROM MyTable t1
    INNER JOIN Mytable t2 ON t1.Email = t2.Email
      AND t1.id <> t2.id
  ) ss
) m;

MySQL频率/分布

1 个答案: