如何计算mysql中的子查询

我有一个疑问：

Select
  DATE_FORMAT(calls.call_datetime, "%Y-%m-%d") as `groupdate`
from
  calls
WHERE calls.job_id =1
group by DATE_FORMAT(calls.call_datetime,
"%Y-%m-%d")

返回：

分组日期

2018-07-06

2018-07-06

有两排。如何对上述查询进行计数，使结果变为一行的“2”

我想要的是：

结果

二,

谢谢，请帮我试试：

Select
  Count(DATE_FORMAT(calls.call_datetime, "%Y-%m-%d")) as `groupdate`
from
  calls
WHERE calls.job_id =1
group by DATE_FORMAT(calls.call_datetime,
"%Y-%m-%d")

您应该使用count（）

---

最简单的方法是使用

Select
  COUNT(*) as Result
from
  calls
WHERE calls.job_id =1
group by DATE_FORMAT(calls.call_datetime, "%Y-%m-%d")

或者是一个交付表方法

SELECT
 COUNT(*) as Result
FROM (
  Select
    DATE_FORMAT(calls.call_datetime, "%Y-%m-%d") as `groupdate`
  from
    calls
  WHERE calls.job_id =1 
  group by DATE_FORMAT(calls.call_datetime, "%Y-%m-%d")
) AS a

---

两个查询应返回您想要的相同预期结果。

问题不明确。因为一个日期有两行，所以期望的答案是“2”吗？如果有多个约会呢

SELECT COUNT(*) FROM calls WHERE ...; -- delivers "2"

SELECT COUNT(*) FROM calls WHERE ...; GROUP BY DATE(datetime) -- also delivers "2"

SELECT COUNT(DISTINCT DATE(call_datetime)) FROM calls WHERE ...; -- delivers "1"

---

可能重复@Alexander-不，这不是同一个问题。没错，这个问题的第二个最佳答案对这个问题来说是正确的，但对那个问题来说是错误的。

SELECT COUNT(*) FROM calls WHERE ...; -- delivers "2"

SELECT COUNT(*) FROM calls WHERE ...; GROUP BY DATE(datetime) -- also delivers "2"

SELECT COUNT(DISTINCT DATE(call_datetime)) FROM calls WHERE ...; -- delivers "1"