php - mysql返回多级数组

Question

我正在使用 PHP 应用程序中相关的三个 MySQL 表。有很多问题的测试，问题有很多答案。我想做的是循环遍历数据，如下所示：

foreach($tests as $test)
{
    echo $test->testName;

    foreach($test->questions as $question)
    {
        echo $question->questionText;

        foreach($question->answers as $answer)
        {
            echo $answer->answerText;
        }
    }
}

我想知道的是 MySQL 查询和 PHP 代码会以这种方式循环遍历它吗？

编辑MySQL 不能返回这样的数组，我应该说的是 MySQL + PHP 代码会是什么样子。

为清楚起见，表格是测试、问题和答案。questions 表包含一个 test_id 列，answers 包含一个 question_id

谢谢！

我希望找回的结构类似于：

array(
    'testName' = 'Test name string',
    'questions' = array(
        array(
            'questionId' = 1,
            'questionText' = 'Question string',
            'answers' = array(
                array(
                    'answerId' = 1,
                    'answerText' = 'Answer string'
                ),
                array(
                    'answerId' = 2,
                    'answerText' = 'Answer string'
                )
            )
        )
    )
);

编辑

我目前的实现如下，我想做的是急切地加载数据而不是执行这么多的查询

$tests = getTests();

foreach($tests as $test){

    $questions = getQuestions($test->id);

    foreach($questions as $question){

        $answers = getAnswers($question->id);

        foreach($answers as $answer){

            // do answer things

        }

    }

}

Answer 1

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");


$query = "SELECT testname, question, answer
    FROM tests 
    JOIN questions ON (tests.id = question.test_id)
    JOIN answers ON (questions.id = answers.id)
    WHERE 'your condition'
    ORDER BY tests.id, question.id"

$result = $mysqli->query($query);

您可以使用 in_array 函数来获得您想要的结果（您可以制作数组或打印结果）。这是打印值的示例。（您可以使用 id 代替 name 来改进）

$tests_arr = array();
$questions_arr = array();

while( $row = $result->fetch_array(MYSQLI_ASSOC) )
{
    if(!in_array($row['testname'], $test_arr)
    {
        $test_arr[] = $row['testname'];
        echo $row['testname'];
    }

    if(!in_array($result['question'], $question_arr))
    {
        $questions_arr[] = $row['question'];
        echo $row['question'];
    }

    echo $row['answer'];
}

Answer 2

MySQL 不能以您想要的格式返回数组结构 - 它可以做的最好的事情是 2D 数组来表示列/行。

两种选择都将在 PHP 中实现。

1. 为每个测试运行一个查询；遍历每个测试并运行查询以获取此测试的所有问题；遍历每个问题并运行查询以获取该问题的答案。

例子：

// get each test
$tests = array();
$results = $mysqli->query('SELECT id, name FROM tests;');
while ($test = $results->fetch_object()) $tests[] = $test;
foreach ($tests as $t => $test) {
    // get all questions for this test
    $tests[$t]['questions'] = array();
    $results = $mysqli->query('SELECT id, name FROM questions WHERE test_id = ' . $test['id']);
    while ($question = $results->fetch_object()) $tests[$t]['questions'][] = $question;
    foreach ($tests[$t]['questions'] as $q => $question) {
        // get all answers for this question
        $tests[$t]['questions'][$q]['answers'] = array();
        $results = $mysqli->query('SELECT id, name FROM answers WHERE question_id = ' . $question['id']);
        while ($answer = $results->fetch_object()) $tests[$t]['questions'][$q][] = $answer;
    }
}

2. 使用 MySQL 连接运行单个查询并预先遍历结果列表以构建您想要的格式的数组。

Answer 3

在纯 SQL 中，您可以进行连接，但这会使您的行变平，行数将是测试 * 问题 * 答案。然后，您必须解析属性并将其映射到适当的对象。像这样的东西：

$sql = "SELECT * FROM tests t, questions q, answers a  WHERE q.t_id = t.id AND a.q_id = q.id";
$rows = // do your fetching, pdo, mysqli, whatever, just get the rows

$tests = array();
foreach($rows as $row) {
    // map test
    if(!isset($tests[$row['t_id']])) $tests[$row['t_id']] = array(
       // set test attributes here (without the questions)
    );

    // map question
    if(!isset($tests[$row['t_id']]['questions'][$row['q_id']])) = array(
       // set question attributes here (without the answers)
    );

    // map answer
    if(!isset($tests[$row['t_id']]['questions'][$row['q_id']]['answers'][$row['a_id']])) = array(
       // set answer attributes here 
    );
}

$tests 数组现在应该具有您需要的层次结构，但一定要对此进行测试，但即使它不起作用，您也会明白这一点。

话虽如此，看看一些 ORM-s。对于一个最简单的（以一种好的方式）看看这个： http: //phpmaster.com/database-interaction-made-easy-with-notorm/

本教程很容易完成，最后你会得到你所要求的，一组具有属性和基于表关系的良好层次结构的对象。

Answer 4

你会LEFT JOIN这样的结果是这样的：

test1|question1|answer1
test1|question1|answer2
test1|question2|answer1
test1|question2|answer2
test2|question1|answer1

然后你以这样一种方式循环遍历它，使它们总是按那个顺序出现，所以编写一个循环来获取一行，如果测试自上一个测试以来发生了变化，则移动到数组中的下一个测试。重复问题，最后添加答案。

一些伪代码

fetcharray()
if test > lasttest
  arraytest++
  arrayquestion = 0
elif question > lastquestion
  arrayquestion++

array[arraytest][arrayquestion][] = fetched array