【发布时间】:2013-10-03 12:25:49
【问题描述】:
我有这些疑问:
SELECT COUNT(*) FROM t_table WHERE color = 'YELLOW';
SELECT COUNT(*) FROM t_table WHERE color = 'BLUE';
SELECT COUNT(*) FROM t_table WHERE color = 'RED';
有没有办法在一个查询中获得这些结果?
【问题讨论】:
我有这些疑问:
SELECT COUNT(*) FROM t_table WHERE color = 'YELLOW';
SELECT COUNT(*) FROM t_table WHERE color = 'BLUE';
SELECT COUNT(*) FROM t_table WHERE color = 'RED';
有没有办法在一个查询中获得这些结果?
【问题讨论】:
SELECT color, COUNT(*) FROM t_table GROUP BY color
【讨论】:
SELECT color, COUNT(*) FROM t_table WHERE color IN ('YELLOW', 'BLUE', 'RED') GROUP BY color
如果您希望结果在一行中,您可以使用:
SELECT
SUM(IF(color = 'YELLOW', 1, 0)) AS YELLOW,
SUM(IF(color = 'BLUE', 1, 0)) AS BLUE,
SUM(IF(color = 'RED', 1, 0)) AS RED
FROM t_table
【讨论】:
SUM(color='YELLOW')等
SUM(IF(color = 'YELLOW', 0.75, 0)) 这样的权重来计算。
SELECT 'yellow' as color ,COUNT(*) FROM t_table WHERE color = 'YELLOW'
union
SELECT 'blue' , COUNT(*) FROM t_table WHERE color = 'BLUE'
union
SELECT 'red',COUNT(*) FROM t_table WHERE color = 'RED';
或
select color, count(*) from table where color in ('red', 'blue', 'yellow') group by 1
【讨论】:
您可以使用子查询来做到这一点。
SELECT(
SELECT COUNT(*) FROM t_table WHERE color = 'YELLOW',
SELECT COUNT(*) FROM t_table WHERE color = 'BLUE',
SELECT COUNT(*) FROM t_table WHERE color = 'RED'
);
【讨论】:
我认为这也适合你
select count(*) as anc,(select count(*) from Patient where sex='F')as
patientF,(select count(*) from Patient where sex='M') as patientM from anc
你甚至可以像这样选择和统计相关的表
select count(*) as anc,(select count(*) from Patient where
Patient.Id=anc.PatientId)as patientF,(select count(*) from Patient where
sex='M') as patientM from anc
【讨论】:
这是我的答案:Este Ejemplo SQL Indica la cantidad de un Grupo y Suma los encontrado con S y N por separado。 Se que no es la Respuesta pero puede ser usado para otros casos。本迪托海以色列。
SELECT sm_med_t_servicios.id as identidad, count(sm_adm_t_admision.id) as cantidad ,
SUM(IF(sm_adm_t_admision.atendido = 'S', 1, 0)) AS atendidos,
SUM(IF(sm_adm_t_admision.atendido = 'N', 1, 0)) AS por_ver
FROM sm_med_t_servicios
LEFT JOIN sm_adm_t_admision ON sm_med_t_servicios.id = sm_adm_t_admision.sm_med_t_servicios_id
WHERE sm_med_t_servicios.m_empresas_id = '2'
GROUP BY sm_med_t_servicios.id
我希望这对你有帮助。
【讨论】: