Question

对于不同条件，我在单列上需要3种不同的计数。

表格结构：

interview-
id-int(10)
c_id-int(10)
experience-varchar2(100)

经验有3个不同的价值 - 1）正 2）负 3）中性

对于c_id = 10的条件，我需要3个不同的“count_positive”，“count_negative”和“count_neutral”计数。

我知道它可以通过3个不同的查询获得。我可以通过单个查询获得3个计数吗？

Answer 1

SELECT
   SUM(CASE experience
          WHEN 'positive' THEN 1
          ELSE 0
       END) AS CountPositive
   , SUM(CASE experience
            WHEN 'negative' THEN 1
            ELSE 0
         END) AS CountNegative
   , SUM(CASE experience
            WHEN 'neutral' THEN 1
            ELSE 0
         END) AS CountNeutral
FROM Interview
WHERE c_id = 10

Answer 2

select 'Positive Count' , count(*)
from interview
where experience = 'positive'
UNION
select 'Negative Count' , count(*)
from interview
where experience = 'negative'
UNION
select 'Neutral' , count(*)
from interview
where experience = 'neutral'

Answer 3

这是Adam Wenger的修改版本答案：

SELECT
   COUNT(CASE experience
          WHEN 'positive' THEN 1
          ELSE NULL
       END) AS CountPositive
   , COUNT(CASE experience
            WHEN 'negative' THEN 1
            ELSE NULL
         END) AS CountNegative
   , COUNT(CASE experience
            WHEN 'neutral' THEN 1
            ELSE NULL
         END) AS CountNeutral
FROM Interview
WHERE c_id = 10

Answer 4

我认为这很好用：

select 'count of ' + experience, 
       count(experience) 
from interview 
where c_id=10
group by experience

Answer 5

我在Codeigniter中获得了有效记录查询的解决方案：

$this->db->select('SUM(CASE experience
          WHEN "positive" THEN 1
          ELSE 0
       END) AS CountPositive
   , SUM(CASE experience
            WHEN "negative" THEN 1
            ELSE 0
         END) AS CountNegative
   , SUM(CASE experience
            WHEN "neutral" THEN 1
            ELSE 0
         END) AS CountNeutral');

$this->db->where('c_id',10);
$query=$this->db->get('interview');
$result=$query->result();

$interview_experience=$result[0];
$positive_count=$interview_experience->CountPositive;
$negative_count=$interview_experience->CountNegative;
$neutral_count=$interview_experience->CountNeutral;

针对不同条件计算单列

5 个答案: