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Php 计算mysql中的出现次数_Php_Mysql_Sql - Fatal编程技术网

Php 计算mysql中的出现次数

Php 计算mysql中的出现次数,php,mysql,sql,Php,Mysql,Sql,我想计算表中“应答”字段的每个值相对于响应的出现次数 我试过了,但没有成功 resp_id visitorID surveyID questionID response answer userID 43 777 163 736 MS 0 1 42 777 163 736 Rohit 1 1 41 777 163

我想计算表中“应答”字段的每个值相对于响应的出现次数

我试过了,但没有成功

resp_id visitorID surveyID questionID response answer userID
     43       777      163        736 MS            0      1
     42       777      163        736 Rohit         1      1
     41       777      163        736 Virat         1      1
     40       776      163        736 MS            1      1
     39       776      163        736 Rohit         3      1
     38       776      163        736 Virat         1      1
     37       775      163        736 MS            0      1 
     36       775      163        736 Rohit         1      1 
     35       775      163        736 Virat         2      1
     34       774      163        736 MS            2      1
     33       774      163        736 Rohit         3      1
     32       774      163        736 Virat         1      1

其中,
$q
等于唯一响应值。

按visitorID使用分组 即


您希望使用count和group by语句获取每种类型答案的编号:

 SELECT count(answer) as answer_cnt FROM `sg_finished_surveys`
 WHERE resopnse = $q group by visiterID
这将计算每个答案的实例数,并给出实际答案

您的
where
子句(resopnse!=response)中也有输入错误


您可能还想查看我发布的这篇文章,它涵盖了这类查询以及更多内容。

$q
需要包含在单个报价中

SELECT 
    count(*) as answer_cnt,
    `answer`
FROM 
    `sg_finished_surveys`
WHERE 
    response = '$q'
GROUP BY 
    `answer`

否,我想计算您在代码中编写的“resopnse”中的答案列中出现的1,2,3,4,5,而不是“response”,使用visitorIDValid组的答案值似乎包括0,1,2和3。那么4和5是从哪里来的呢?
SELECT 
    count(*) as answer_cnt,
    `answer`
FROM 
    `sg_finished_surveys`
WHERE 
    response = '$q'
GROUP BY 
    `answer`
SELECT count(*) as answer_cnt
FROM `sg_finished_surveys`
WHERE resopnse = '$q' GROUP BY `answer`
SELECT COUNT(answer)
FROM `sg_finished_surveys`
WHERE respondence = '".$q."'
GROUP BY answer