0

我目前正在使用以下来计算多列，它基本上将 answer1、answer2 .....etc 等列的总数增加到 answer30。

所以结果看起来像

1x
4x
3x
3x
4x

ETC

有没有更好的方法可以将所有这些代码放入一个查询中？

    $sql1 = <<<SQL
        SELECT answer1, COUNT(answer1)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer1
    SQL;
    if(!$result1 = $db->query($sql1)){ die('There was an error running the query [' . $db->error . ']');}

    while($row1 = $result1->fetch_assoc()){ 
    echo $row1['COUNT(answer1)'] . ' X <strong>' . $answer1 . '</strong><br />';
    }

    $sql2 = <<<SQL
        SELECT answer2, COUNT(answer2)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer2
    SQL;
    if(!$result2 = $db->query($sql2)){ die('There was an error running the query [' . $db->error . ']');}

    while($row2 = $result2->fetch_assoc()){ 
    echo $row2['COUNT(answer2)'] . ' X <strong>' . $answer2 . '</strong><br />';
    }

    $sql3 = <<<SQL
        SELECT answer3, COUNT(answer3)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer3
    SQL;
    if(!$result3 = $db->query($sql3)){ die('There was an error running the query [' . $db->error . ']');}

    while($row3 = $result3->fetch_assoc()){ 
    echo $row3['COUNT(answer3)'] . ' X <strong>' . $answer3 . '</strong><br />';
    }

    $sql4 = <<<SQL
        SELECT answer4, COUNT(answer4)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer4
    SQL;
    if(!$result4 = $db->query($sql4)){ die('There was an error running the query [' . $db->error . ']');}

    while($row4 = $result4->fetch_assoc()){ 
    echo $row4['COUNT(answer4)'] . ' X <strong>' . $answer4 . '</strong><br />';
    }

    $sql5 = <<<SQL
        SELECT answer5, COUNT(answer5)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer5
    SQL;
    if(!$result5 = $db->query($sql5)){ die('There was an error running the query [' . $db->error . ']');}

    while($row5 = $result5->fetch_assoc()){ 
    echo $row5['COUNT(answer5)'] . ' X <strong>' . $answer5 . '</strong><br />';
    }

    $sql6 = <<<SQL
        SELECT answer6, COUNT(answer6)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer6
    SQL;
    if(!$result6 = $db->query($sql6)){ die('There was an error running the query [' . $db->error . ']');}

    while($row6 = $result6->fetch_assoc()){ 
    echo $row6['COUNT(answer6)'] . ' X <strong>' . $answer6 . '</strong><br />';
    }

    $sql7 = <<<SQL
        SELECT answer7, COUNT(answer7)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer7
    SQL;
    if(!$result7 = $db->query($sql7)){ die('There was an error running the query [' . $db->error . ']');}

    while($row7 = $result7->fetch_assoc()){ 
    echo $row7['COUNT(answer7)'] . ' X <strong>' . $answer7 . '</strong><br />';
    }

    $sql8 = <<<SQL
        SELECT answer8, COUNT(answer8)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer8
    SQL;
    if(!$result8 = $db->query($sql8)){ die('There was an error running the query [' . $db->error . ']');}

    while($row8 = $result8->fetch_assoc()){ 
    echo $row8['COUNT(answer8)'] . ' X <strong>' . $answer8 . '</strong><br />';
    }

    $sql9 = <<<SQL
        SELECT answer9, COUNT(answer9)
        FROM `QuestionnaireAnswers`
        WHERE questionnaireID='$questionnaireID'
        GROUP BY answer9
    SQL;
    if(!$result9 = $db->query($sql9)){ die('There was an error running the query [' . $db->error . ']');}

    while($row9 = $result9->fetch_assoc()){ 
    echo $row9['COUNT(answer9)'] . ' X <strong>' . $answer9 . '</strong><br />';
    }
4

1 回答 1

1

一种方法：

select answerNum, answerVal, count(answerVal) from
(select n.num answerNum,
        case n.num
            when 1 then a.answer1
            when 2 then a.answer2
            ...
        end answerVal
 from (select 1 num union select 2 union ...) n
 cross join `QuestionnaireAnswers` a
 WHERE questionnaireID='$questionnaireID') sq
group by answerNum, answerVal
于 2013-06-20T11:07:06.283 回答