您应该使用count（）

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

这个问题模棱两可。因为一个日期有两行，所以期望的答案是“2”吗？如果不止一次约会呢？

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"

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")

或者是一个传递表的方法

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

两个查询都应返回所需的相同预期结果。

尝试：

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")