MYSQL/PHP统计重复问题的答案

我有点困惑我该怎么做。但我会尽力解释清楚

我正在创建一个类似以下内容的报告：

但还有很多问题

所以有很多人回答了这项调查，所有的记录都在数据库中，我需要用它来制作这份报告

我创建了一个查询，该查询从连接多个表中提取所有相关数据，如下所示：

表格 -调查 -调查入口 -调查问题 -调查 -硬件服务

        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'

好的，不，只是为了证明这是一张我正在提取的数据的图片。这项调查共有30个问题，但在这里，我只为每个条目显示了4行

因此，它们是相同的问题等，但不同的条目

现在我该怎么做呢

然后得到每个问题中表示“是”的数字和表示“否”的数字，这样我就可以用php生成这个

---

如果您需要更多信息，请告诉我。

对于性别报告，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`

对于年龄组报告，与上面类似，只需更改

调查问题

ID

。所以代码是这样的

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`

---

由于您拥有所有数据，我们可以使用

分组方式

，查看每个问题所有唯一答案的计数。为此，我们可以使用以下方法：

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

然后，它将为您提供每个问题（及其id），以及该问题的唯一答案和唯一答案的计数

或者在您提供的一个选项中完成所有操作：

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`

---

其中，我在初始选择中添加了

count（*）

，并在末尾添加了

GROUP BY surveyQuestions.ID，survey\u meta.answer

，将

count（*）作为总数添加到您的
select
中，然后添加
GROUP BY`surveyQuestions`.`ID`，“调查”.“回答”
@Sean你能帮我把这个写进一个答案里吗？我不知道该怎么做。你好，谢谢你的回答。然而，问题远不止性别和年龄组一，我只记下这两个问题，因为它们是不同的选择。所以每个问题要么是答案1要么是答案2，年龄范围是唯一有5个答案可供选择的类型。我该如何处理这个问题？谢谢你，我想这正是我在mysql方面需要的。现在只需要对php执行此操作：），顺便说一句，总数后缺少一个逗号。事实上，即使答案为0，也可以对其进行编辑以显示带有答案的行吗？例如，对于种族，我有5个可能的答案，在本例中，如果选择了4个答案，那么在表中它只显示4个，但我需要它显示5个，并在选项旁边放一个0