计算使用where的行数

Question

m mysql中的新内容

这是我的表

现在我想计算＆＃34; count_id＆＃34;其中＆＃39; questionID＆＃39;大于2

Answer 1

试试这个：

SELECT COUNT(count_id) FROM myTable WHERE questionID > 2

Answer 2

select count(Count_ID),QuestionID,SurveyId from table 
where QuestionID>2
group by QuestionID,SurveyID

Answer 3

您也可以尝试以下声明：

select count(count_id) CountOfID,count_id from mytable 
where questionID > 2 group by count_id;

Answer 4

select  count(count_id) from yourtable where questionID > 2

Answer 5

如果您想要计算唯一ID：

select count(DISTINCT count_id) from table_name where questionID > 2

Answer 6

SELECT COUNT（count_id）FROM table_name WHERE questionID＆gt; 2

Answer 7

按Count_ID分组并计算他们不同的问题。留在那些有两个以上的人。然后计算你得到多少个ID。

select count(*)
from
(
  select count_id
  from mytable
  group by count_id
  having count(distinct questionid) > 2
) x;

编辑：如果count_id + questionid恰好是该表的唯一，则可以将count(distinct questionid)替换为count(*)。