mysql - 如何让这 2 个 SQL 查询返回 1 个结果集？（他们选择相同的项目，使用不同的 where 子句）

Question

我想将此查询合并为一个，以按日期返回 2 个值（“scount_atp”，其中 estado=1 和“scount_xxx”，其中 estado=4）：

SELECT DATE_FORMAT(FROM_UNIXTIME(fecha),"%m-%d-%Y") as sdate, COUNT(id) as scount_atp FROM entradas WHERE estado = '1' AND fecha <= UNIX_TIMESTAMP(NOW()) AND fecha >= UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 1 YEAR)) GROUP BY sdate

SELECT DATE_FORMAT(FROM_UNIXTIME(fecha),"%m-%d-%Y") as sdate, COUNT(id) as scount_xxx FROM entradas WHERE estado = '4' AND fecha <= UNIX_TIMESTAMP(NOW()) AND fecha >= UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 1 YEAR)) GROUP BY sdate

我是 MySQL 的新手，所以如果这太基本了，我很抱歉......

谢谢大家的帮助！

Answer 1

SELECT 
  DATE_FORMAT(FROM_UNIXTIME(fecha),"%m-%d-%Y") as sdate, 
  SUM(IF(estado = '1',1,0)) as scount_atp,
  SUM(IF(estado = '4',1,0)) as scount_xxx
FROM entradas 
WHERE estado IN('1','4')
  AND fecha <= UNIX_TIMESTAMP(NOW()) 
  AND fecha >= UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 1 YEAR)) 
GROUP BY sdate

让我在这里添加一个不相关的东西：如果该estado列是 type INT，您可能想要编写1而不是'1'- 或者您可能以全表扫描结束，将 int 列转换为字符串以进行比较。

Answer 2

我过去做过类似的事情，这对我有用：

SELECT 
   DATE_FORMAT(FROM_UNIXTIME(fecha),"%m-%d-%Y") as sdate, 
   COUNT(IF(estado = '1', 1, NULL)) as scount_atp, 
   COUNT(IF(estado = '4', 1, NULL)) as scount_xxx 
FROM entradas 
WHERE 
   fecha <= UNIX_TIMESTAMP(NOW()) 
   AND 
   fecha >= UNIX_TIMESTAMP(DATE_SUB(NOW(), INTERVAL 1 YEAR))
GROUP BY sdate