在SQL Server中需要帮助以列出和计算动物的状态

您也可以使用 PIVOT

DECLARE @T TABLE(name VARCHAR(10),   status INT)
INSERT INTO @T
VALUES
('frog',  1 ),
('frog',  0 ),
('frog',  0 ),
('rabbit',  0 ),
('rabbit',  1 ),
('rabbit',  0 ),
('rabbit',  0 )

SELECT name, [0] status0, [1] status1 FROM @T
    PIVOT(COUNT(status) FOR status IN ([0],[1])) PVT

结果：

name       status0     status1
---------- ----------- -----------
frog       2           1
rabbit     3           1