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I'm new to PHP and MySQL and am struggling with this... I am collecting form data from a PHP insert page. It is a survey questionnaire. There are 20 questions, and all the answers will either be A, B and C. The table has an id column, and a column for each question (Q1, Q2, Q3... Q20). The data may look something like this;

+------+-----+
|  id  |  1  |
+------+-----+
|  Q1  |  A  |
|  Q2  |  B  |
|  Q3  |  A  |
|  .   |  .  |
|  .   |  .  |
|  .   |  .  |
|  Q20 |  C  |
+------+-----+

Now, what I am trying to do is to count how many values of A, B and C appear in a single row (with id=1 for example).

I've found lots of ways of counting values in multiple columns, but so far have not found a way to count/group values in a single row.

4

3 回答 3

0

在 PHP 中，您可以将查询结果加载到一个数组中，然后用于array_count_values获取每个答案的计数：

$array = array();

while ($row = $result->fetch_assoc()) {
    $array[] = $row;
}

// print_r(array_count_values($array); 

foreach(array_count_values($array) as $key => $value)
{
     echo 'Answer ' . $key . ' was chosen ' . $value . ' times <br>';
}

工作小提琴

于 2013-04-18T14:35:26.417 回答
0

免责声明：根据记忆，因此可能不完全有效。

这个怎么样？

SELECT 1 AS 1, (SELECT COUNT(1) FROM Questions WHERE 1 = 'A') AS A, (SELECT COUNT(1) FROM Questions WHERE 1 = 'B') AS B, (SELECT COUNT(1) FROM Questions WHERE 1 = 'C') 作为 C

希望这可以帮助。

于 2013-04-18T14:36:43.987 回答
0

以下是使用 MySQL 查询的方法：

select id,
  sum(case when `1` = 'A' then 1 else 0 end) as CountA,
  sum(case when `1` = 'B' then 1 else 0 end) as CountB,
  sum(case when `1` = 'C' then 1 else 0 end) as CountC
from SurveyTable
group by id
order by id;

这是一个测试数据有限的SQL Fiddle。

附录。卡洛斯发布了一个更新的结构，导致以下答案。希望这些很接近:)

这将为您提供一排非常宽的总计：

select
  sum(case when Q1 = 'A' then 1 else 0 end) as Q1CountA,
  sum(case when Q1 = 'B' then 1 else 0 end) as Q1CountB,
  sum(case when Q1 = 'C' then 1 else 0 end) as Q1CountC,
  sum(case when Q2 = 'A' then 1 else 0 end) as Q2CountA,
  sum(case when Q2 = 'B' then 1 else 0 end) as Q2CountB,
  sum(case when Q2 = 'C' then 1 else 0 end) as Q2CountC,
  sum(case when Q3 = 'A' then 1 else 0 end) as Q3CountA,
  sum(case when Q3 = 'B' then 1 else 0 end) as Q3CountB,
  sum(case when Q3 = 'C' then 1 else 0 end) as Q3CountC,
  sum(case when Q4 = 'A' then 1 else 0 end) as Q4CountA,
  sum(case when Q4 = 'B' then 1 else 0 end) as Q4CountB,
  sum(case when Q4 = 'C' then 1 else 0 end) as Q4CountC,
  sum(case when Q5 = 'A' then 1 else 0 end) as Q5CountA,
  sum(case when Q5 = 'B' then 1 else 0 end) as Q5CountB,
  sum(case when Q5 = 'C' then 1 else 0 end) as Q5CountC
from SurveyTable;

如果您想每个问题获得一行，请尝试以下操作：

select
  QuestionID,
  sum(case when Answer = 'A' then 1 else 0 end) as CountA,
  sum(case when Answer = 'B' then 1 else 0 end) as CountB,
  sum(case when Answer = 'C' then 1 else 0 end) as CountC
from (
    select 'Question1' as QuestionID, Q1 as Answer from surveytable
    union all select 'Question2', Q2 from surveytable
    union all select 'Question3', Q3 from surveytable
    union all select 'Question4', Q4 from surveytable
    union all select 'Question5', Q5 from surveytable) x
group by QuestionID

这里有一个小提琴。

另一个附录： 需要的计数ID，并且因为ID每行有一个，所以不需要SUM.

这改变了方法。它首先将答案串在一起：

concat(q1,q2,q3,q4,q5) -- result for ID=1 in the test data: 'ABCAC'

A...然后它会从字符串中吸取每一次出现：

replace(concat(q1,q2,q3,q4,q5), 'A', '') -- result for ID=1: 'BCC'

...第一个字符串 ( ABCAC) 是长度5，第二个字符串 ( BCC) 是长度 3。长度的差异是A答案的数量：2。这差不多就是我能解释的了。现在进行查询：

select
  id,
  5 - length(replace(concat(q1,q2,q3,q4,q5), 'A', '')) AS CountA,
  5 - length(replace(concat(q1,q2,q3,q4,q5), 'B', '')) AS CountB,
  5 - length(replace(concat(q1,q2,q3,q4,q5), 'C', '')) AS CountC
from surveytable;

更新的 Fiddle 在这里。

这仅提供原始数据。格式化会有点棘手，但应该不会太糟糕，特别是如果您使用前端语言进行格式化。如果您必须为此使用 MySQL，将上述内容放入子查询并在外部查询中应用格式可能会更容易：

select
  id,
  CONCAT('You have chosen ' ...and miles of formatting logic using CountA, etc)
from (
  select
    id,
    5 - length(replace(concat(q1,q2,q3,q4,q5), 'A', '')) AS CountA,
    5 - length(replace(concat(q1,q2,q3,q4,q5), 'B', '')) AS CountB,
    5 - length(replace(concat(q1,q2,q3,q4,q5), 'C', '')) AS CountC
  from surveytable) x
于 2013-04-18T14:46:01.973 回答