Question

嗨，我在下面有这个SQL

数据

CREATE TABLE data
    (id int, name varchar(9), room varchar(9), paid varchar(9));

INSERT INTO data
    (id, name, room, paid)
VALUES
    (1, 'jane', 'room1','0'),
    (2, 'any', 'room2', '1'),
    (3, 'mandy', 'room2', '1'),
    (4, 'vivian', 'room2', '1'),
    (5, 'amber', 'room3', '1'),
    (6, 'fany', 'room4', '1'),
    (7, 'diana', 'room4', '0');

查询

SELECT 
  data.room, COUNT(data.id) as total
FROM 
  data
WHERE 
  data.paid = 0
GROUP BY
  data.room

示例链接sqlfiddle

现在的结果是

room 1 = 1 
room 4 = 1

我想要的结果

room 1 = 1
room 2 = 0
room 3 = 0
room 4 = 1

任何人都可以帮助我，谢谢

Answer 1

SELECT 
  data.room, SUM(data.paid = 0) as total
FROM 
  data
GROUP BY
  data.room;

查看Demo on SQL Fiddle。

Answer 2

SELECT 
  data.room, SUM(CASE WHEN data.paid = '0' THEN 1 ELSE 0 END) as total
FROM 
  data
GROUP BY
  data.room

请参阅SQLFiddle

Answer 3

cdaiga的答案是正确的，但依赖于您的值（0和1允许您执行SUM技巧。一般情况下，您希望将原始表格与WHERE和GROUP BY生成的表格保持一致。

SELECT DISTINCT original.room, IFNULL(pr.total, 0) AS total
FROM data original LEFT OUTER JOIN (
    SELECT 
      data.room as paidrooms, COUNT(data.id) as total
    FROM 
      data
    WHERE
      data.paid = 0
    GROUP BY
      data.room
) as pr on original.room = pr.paidrooms

链接到SQLFiddle

Answer 4

据我所知，您希望列出所有房间，每个房间都有未付账单。你冷使用嵌套的选择如下：

    SELECT 
    d1.room, 
    (SELECT COUNT(d2.room) FROM data d2 WHERE d1.room = d2.room and d2.paid = 0)   as total
    FROM 
        data d1
    GROUP BY
        d1.room

Sql Fiddle here

Mysql查询显示组数据

4 个答案: