Question

基本上，我有三张桌子。我有一个项目表，一个问题表和一个答案表。一个项目可以有很多问题，一个问题可以有很多答案。在问题ID上使用PDO和LEFT JOIN，如何将带有答案的注释转换为多维数组，以便结构如下所示：

[Question] => Array
    (
        [id] => 1
        [question] => 'Random Question'
        [askedBy] => 123
        [answer] => Array
            (
              [0] => Array
                 (
                   [id] => 1
                   [answer] => 'An Answer'
                   [answeredBy] => 123
                 )
               [1] => Array
                 (
                   [id] => 1
                   [answer] => 'Another Answer'
                   [answeredBy] => 123
                 )
            )
      )

最终代码（返回我想要的内容）

$questions = array();
$questionCounter = 0;
$questionID = NULL;


$STH = $DBH->query("SELECT `project_question`.`id` AS question_id, `project_question`.`question`, `project_question`.`userID` AS askedBy, 
                               `project_question`.`created` AS question_created, `project_answer`.`id` AS answer_id, 
                               `project_answer`.`answer`, `project_answer`.`userID` AS answeredBy, 
                               `project_answer`.`accepted`, `project_answer`.`created` AS answer_created
                          FROM `project_question`
                     LEFT JOIN `project_answer`
                            ON `project_question`.`id` = `project_answer`.`questionID`
                         WHERE `project_question`.`projectID` = $args[0]
                           AND `project_question`.`projectPhase` = 2");

    while($row = $STH->fetch(PDO::FETCH_ASSOC)){
      if($row['question_id'] !== $questionID){
        $questions[$questionCounter] = array(
            'id' => $row['question_id'],
            'question' => $row['question'],
            'userID' => $row['askedBy'],
            'created' => $row['question_created'],
            'answers' => array()                  
        );
        array_push($questions[$questionCounter]['answers'], 
             array(
              'id' => $row['answer_id'],
              'answer' => $row['answer'],
              'userID' => $row['answeredBy'],
              'accepted' => $row['accepted'],
              'created' => $row['answer_created']
        ));
        $questionCounter++;
        $questionID = $row['question_id'];
      } else {
        array_push($questions[$questionCounter - 1]['answers'], 
             array(
              'id' => $row['answer_id'],
              'answer' => $row['answer'],
              'userID' => $row['answeredBy'],
              'accepted' => $row['accepted'],
              'created' => $row['answer_created']
        ));
      }          
    }

Answer 1

在一个查询中，您可以请求所有答案，加入问题并将项目加入问题。

创建空数组$projects = array()
遍历每一行foreach ($rows as $row)
将项目及其数据添加到数组中（如果它已经不存在）（使用项目ID作为键）if (!isset($projects[$row->project_id])) { $projects[$row->project_id] = array('col1' => $row->col1,'questions' => array()); }
将问题及其数据添加到项目问题数组中（如果它尚不存在（使用问题ID作为键）if (!isset($projects[$row->project_id]['questions'][$row->question_id])) { $projects[$row->project_id]['questions'][$row->question_id] = array('col2' => $row->col2,'answers' => array()); }
将其数据的答案添加到项目问题答案数组（使用答案ID作为键）$projects[$row->project_id]['questions'][$row->question_id]['answers'][$row->answer_id] = array('col3' => $row->col3);

但是正如您所理解的那样，您将在该查询中获得许多无用的信息。您可以使用一个查询，仅获取项目，遍历它们，在每个周期中将数据添加到数组并再次查询以获取具有特定project_id的问题，迭代它们，在每个周期中将数据添加到数组并再次查询以获得答案具有特定的question_id，迭代它们，并将数据添加到数组。

但是如果我重新考虑这个问题，我认为MySQL的运行速度比PHP快，即使返回一堆无用的数据（我的意思是，重复关于项目和问题的数据）和1个查询与来自一个客户端的50个查询相比要好得多，所以我建议更好地使用第一种方法。

如果没有关于数据库表和列的更多信息，除了伪代码算法之外，你不会得到任何其他信息。

编辑：可能的MySQL选择查询：

SELECT 
    a.id answer_id, a.answer, a.answeredBy, 
    q.id question_id, q.question, q.askedBy, 
    p.id project_id, p.title
FROM answer a
LEFT JOIN question q ON q.id = a.question_id
LEFT JOIN project p ON p.id = q.project_id

表结构

项目 - ＆gt; | id | title |
问题 - ＆gt; | id |问题|问亲project_id |
回答 - ＆gt; | id |回答|回答了question_id |

PHP PDO LEFT JOIN多维数组

1 个答案: