SQL在多列

Question

如果我有一个包含名字，姓氏（复合唯一）和外卖选择的表格，例如

+------+-------+---------+
|First | Last  | Food    |
+------+-------+---------+
|Bob   | Smith | Pizza   |
|Bob   | Smith | Chips   |
|Jim   | Smith | Pizza   |
|Lisa  | Jones | Pizza   |
|Lisa  | Jones | Chinese |
|Lisa  | James | Mexican |
|Eric  | White | Chinese |
|Eric  | White | Chips   |
+------+-------+---------+

我想要的所有行都对应于至少有两种食物偏好的人，其中一种是披萨。即

的结果

+------+-------+---------+
|First | Last  | Food    |
+------+-------+---------+
|Bob   | Smith | Pizza   |
|Bob   | Smith | Chips   |
|Lisa  | Jones | Pizza   |
|Lisa  | Jones | Chinese |
+------+-------+---------+

我已尝试过加入...群组......有计数（*）＆gt; 1但它并不是我想要的工作。

感谢任何帮助，谢谢。

Answer 1

您可以首先获取出现多次的名字和姓氏列表：

SELECT first, last
FROM myTable
GROUP BY first, last
HAVING COUNT(*) > 1;

完成后，您可以将其加入原始表格，并确保只选择这些名称：

SELECT m.*
FROM myTable m
JOIN(
   SELECT first, last
   FROM myTable
   GROUP BY first, last
   HAVING COUNT(*) > 1) tmp ON tmp.first = m.first AND tmp.last = m.last;

但是，这个问题的问题在于，它并没有验证人们首选的食物中至少有一种是披萨。为此，我们可以在第一个子查询的HAVING子句中添加一个附加条件，以检查一行披萨。我是使用条件聚合完成的：

SELECT m.*
FROM myTable m
JOIN(
   SELECT first, last
   FROM myTable
   GROUP BY first, last
   HAVING COUNT(*) > 1 AND SUM(food = 'Pizza') = 1) tmp ON tmp.first = m.first AND tmp.last = m.last;

以下是SQL Fiddle示例。

Answer 2

另一种方式：SQL Server 2012中的P. +。

SELECT FIRST,LAST,FOOD FROM(
SELECT A.FIRST,A.LAST,B.FOOD,
COUNT(*) OVER (PARTITION BY A.FIRST,A.LAST ORDER BY A.FIRST,A.LAST DESC) AS 'POS'
FROM TEST_FOOD A
INNER JOIN TEST_FOOD B
ON A.FIRST = B.FIRST
AND A.LAST = B.LAST
WHERE A.FOOD = 'PIZZA') TB WHERE POS = 2

在SQL Server 2005+中另一种更标准的方式，第一种方法更快，但受MSSQL版本的限制。

SELECT C.FIRST,C.LAST,C.FOOD FROM 
(
SELECT A.FIRST,A.LAST FROM TEST_FOOD A
INNER JOIN TEST_FOOD B
ON A.FIRST = B.FIRST
AND A.LAST = B.LAST
WHERE B.FOOD = 'PIZZA'
GROUP BY A.FIRST,A.LAST
HAVING COUNT(*) = 2
) TB INNER JOIN 
TEST_FOOD C
ON TB.FIRST = C.FIRST
AND TB.LAST = C.LAST

Answer 3

{{1}}

有很多方法可以做到这一点。这只是一种方式。

Answer 4

select first,last,GROUP_CONCAT(food separator ',') 
from people 
group by first,last 
having count(food) > 1 AND SUM(food = 'Pizza') = 1

小组concat将用逗号分隔的字符串带回食物。

+------+-------+---------------+
|First | Last  | Food          |
+------+-------+---------------+
|Bob   | Smith | Pizza,Chips   |
|Lisa  | Jones | Pizza,Chinese |
+------+-------+---------------+

Answer 5

或类似的东西（使用McAdam331＆＃39;小提琴）...

 SELECT DISTINCT y.*
            FROM mytable x
            JOIN mytable y
              ON y.first = x.first
             AND y.last = x.last
            JOIN mytable z
              ON z.first = x.first
             AND z.last = x.last
             AND z.food <> y.food
           WHERE x.food = 'pizza';