MySQL不会带回数据

时间:2012-11-27 14:08:14

标签: mysql null rows

我有以下代码:

SELECT q1, COUNT(q1) as Responses, 
(SELECT COUNT(q1) FROM `acme` WHERE q1 >= 1) AS Total, 
(COUNT(q1) / (SELECT COUNT(q1) FROM `acme` WHERE q1 >= 1 )) *100 AS Percentage
FROM `acme` WHERE q1  >= 1 GROUP BY q1 DESC

很简单,q1字段中只能有四个响应 - 它们是1,2,3或4

见下面的输出:

q1  Responses   Total   percentage
4   109         362     30.1105
3   224         362     61.8785
2   25          362     6.9061
1   4           362     1.1050

现在,如果选项1,2,3或4在q1中,这可以解决,但我遇到的问题是如果Q1中没有4's - 代码将返回行3,2和1而不是4(我猜其实是正确的)。

所以,如果没有4个,我需要它来制作这样的东西......

q1  Responses   Total   percentage
4   0           253     0
3   224         253     88.53754
2   25          253     9.88142
1   4           253     1.58102

希望这是有道理的。

1 个答案:

答案 0 :(得分:1)

试试这个:

select * from (SELECT q1, COUNT(q1) as Responses, 
(SELECT COUNT(q1) FROM `acme` WHERE q1 >= 1) AS Total, 
(COUNT(q1) / (SELECT COUNT(q1) FROM `acme` WHERE q1 >= 1 )) *100 AS Percentage
FROM `acme` WHERE q1  >= 1 GROUP BY q1 DESC
union
select 1,0,0,0 from dual
union
select 2,0,0,0 from dual
union
select 3,0,0,0 from dual
union
select 4,0,0,0 from dual) as t group by q1

http://www.sqlfiddle.com/#!2/84b31/6

或者您可以在没有DUAL

的情况下尝试
select * from (SELECT q1, COUNT(q1) as Responses, 
(SELECT COUNT(q1) FROM `acme` WHERE q1 >= 1) AS Total, 
(COUNT(q1) / (SELECT COUNT(q1) FROM `acme` WHERE q1 >= 1 )) *100 AS Percentage
FROM `acme` WHERE q1  >= 1 GROUP BY q1 DESC
union
select 1,0,0,0
union
select 2,0,0,0
union
select 3,0,0,0
union
select 4,0,0,0) as t group by q1;

http://www.sqlfiddle.com/#!2/97031/19