Question

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？

Answer 1

SELECT color, COUNT(*) FROM t_table GROUP BY color

Answer 2

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

Working example

Answer 3

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

Answer 4

您可以使用子查询执行此操作。

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'
);

Answer 5

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

我希望这会对你有所帮助。

Answer 6

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

您甚至可以选择和统计相关的表格

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

多个COUNT（）用于一个查询中的多个条件（MySQL）

6 个答案: