MYSQL/PHP 计算重复问题的答案

Question

我什至不知道该怎么做。但我会尽力解释。

我正在创建一个如下所示的报告：

但是还有很多问题。

所以有很多人回答了这个调查，所有记录都在数据库中，我需要使用它来制作这份报告。

我创建了查询，从连接多个表中提取所有相关数据，如下所示：

表格 -民意调查 -调查条目 -调查问题 -survey_meta -hw_services

        SELECT
            `surveyEntries`.`ID` AS EntryID,
            `surveyEntries`.`created` AS EntryDate,
            `hw_services`.`name` AS Provider,
            `surveyQuestions`.`ID` AS QuestionID,
            `surveyQuestions`.`label` AS Question,
            `survey_meta`.`answer` AS Answer,
          `surveyQuestions`.`parentID` AS ParentQuestion
        FROM `survey`
        JOIN `surveyQuestions`
            ON `survey`.`ID` = `surveyQuestions`.`surveyID`
        JOIN `surveyEntries`
            ON `survey`.`ID` = `surveyEntries`.`surveyID`
        JOIN `survey_meta`
             ON (`surveyEntries`.`ID` = `survey_meta`.`entryID` AND `surveyQuestions`.`ID` = `survey_meta`.`questionID`)
        JOIN `hw_services`
            ON `surveyEntries`.`hw_serviceID` = `hw_services`.`ID`
        WHERE `hw_services`.`healthwatchID` = '1'
        AND `survey`.`ID` = '1'
        AND `surveyQuestions`.`type` IN ('radio', 'dropdown')
        AND `hw_services`.`ID` = '1697'

好吧，不只是为了证明这是我正在撤回的数据的图片。调查中共有 30 个问题，但这里我只显示每个条目的 4 行。

所以它们是相同的问题等，但条目不同。

现在我该怎么做？

并在每个问题中得到回答是的数字和回答不是的数字，这样我就可以在 php?

中生成这个

如果您需要任何进一步的信息，请告诉我。

Answer 1

对于 GENDER 报告，您可以使用此 sql

SELECT
        `survey_meta`.`answer` AS Gender,
        COUNT(*) AS 'Total Answer'
    FROM `survey`
    JOIN `surveyQuestions`
        ON `survey`.`ID` = `surveyQuestions`.`surveyID`
    JOIN `surveyEntries`
        ON `survey`.`ID` = `surveyEntries`.`surveyID`
    JOIN `survey_meta`
         ON (`surveyEntries`.`ID` = `survey_meta`.`entryID` AND `surveyQuestions`.`ID` = `survey_meta`.`questionID`)
    JOIN `hw_services`
        ON `surveyEntries`.`hw_serviceID` = `hw_services`.`ID`
    WHERE `hw_services`.`healthwatchID` = '1'
    AND `survey`.`ID` = '1'
    AND `surveyQuestions`.`type` IN ('radio', 'dropdown')
    AND `hw_services`.`ID` = '1697'
    AND `surveyQuestions`.`ID` = 9
GROUP BY `survey_meta`.`answer`

对于AGE GROUP报告，与上面类似，只需更改surveyQuestions.ID。所以代码会像这样

SELECT
        `survey_meta`.`answer` AS Gender,
        COUNT(*) AS 'Total Answer'
    FROM `survey`
    JOIN `surveyQuestions`
        ON `survey`.`ID` = `surveyQuestions`.`surveyID`
    JOIN `surveyEntries`
        ON `survey`.`ID` = `surveyEntries`.`surveyID`
    JOIN `survey_meta`
         ON (`surveyEntries`.`ID` = `survey_meta`.`entryID` AND `surveyQuestions`.`ID` = `survey_meta`.`questionID`)
    JOIN `hw_services`
        ON `surveyEntries`.`hw_serviceID` = `hw_services`.`ID`
    WHERE `hw_services`.`healthwatchID` = '1'
    AND `survey`.`ID` = '1'
    AND `surveyQuestions`.`type` IN ('radio', 'dropdown')
    AND `hw_services`.`ID` = '1697'
    AND `surveyQuestions`.`ID` = 10
GROUP BY `survey_meta`.`answer`

Answer 2

由于您拥有所有数据，我们可以使用 group by 来查看每个问题的所有唯一答案的计数。为此，我们可以使用如下内容：

SELECT QuestionId, Question, Answer, count(*) 
FROM (PUT YOUR SELECT HERE) 
GROUP BY QuestionId, Answer

然后将为您提供每个问题（及其 ID），以及该问题的唯一答案和该唯一答案的计数。

或者在您提供的 select 中全部完成：

SELECT
            `surveyEntries`.`ID` AS EntryID,
            `surveyEntries`.`created` AS EntryDate,
            `hw_services`.`name` AS Provider,
            `surveyQuestions`.`ID` AS QuestionID,
            `surveyQuestions`.`label` AS Question,
            `survey_meta`.`answer` AS Answer,
             count(*) as Total,
            `surveyQuestions`.`parentID` AS ParentQuestion
        FROM `survey`
        JOIN `surveyQuestions`
            ON `survey`.`ID` = `surveyQuestions`.`surveyID`
        JOIN `surveyEntries`
            ON `survey`.`ID` = `surveyEntries`.`surveyID`
        JOIN `survey_meta`
             ON (`surveyEntries`.`ID` = `survey_meta`.`entryID` AND `surveyQuestions`.`ID` = `survey_meta`.`questionID`)
        JOIN `hw_services`
            ON `surveyEntries`.`hw_serviceID` = `hw_services`.`ID`
        WHERE `hw_services`.`healthwatchID` = '1'
        AND `survey`.`ID` = '1'
        AND `surveyQuestions`.`type` IN ('radio', 'dropdown')
        AND `hw_services`.`ID` = '1697'
        GROUP BY `surveyQuestions`.`ID`, `survey_meta`.`answer`

我在开头的 select 和末尾的 GROUP BY surveyQuestions.ID, survey_meta.answer 中添加了 count(*)

MYSQL/PHP 计算重复问题的答案

MYSQL/PHP count answers to reoccurring questions

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