使用 PHP 多维数组将 MySQL 转换为 JSON

【问题标题】：Using PHP multidimensional arrays to convert MySQL to JSON使用 PHP 多维数组将 MySQL 转换为 JSON
【发布时间】：2014-08-23 12:08:18
【问题描述】：

Here's my table structure.

我正在尝试将 MySQL 转换为嵌套 JSON，但无法弄清楚如何在 PHP 中构建多维数组。

我想要的结果是这样的：

[
{
    "school_name": "School's Name",
    "terms": [
        {                                       
            "term_name":"FALL 2013",
            "departments": [
                {
                    "department_name":"MANAGEMENT INFO SYSTEMS",
                    "department_code":"MIS",
                    "courses": [
                        {
                            "course_code":"3343",
                            "course_name":"ADVANCED SPREADSHEET APPLICATIONS",
                            "sections": [
                                {
                                    "section_code":"18038",
                                    "unique_id": "mx00fdskljdsfkl"
                                },
                                {
                                    "section_code":"18037",
                                    "unique_id": "mxsajkldfk57"
                                }
                            ]
                        },
                        {
                            "course_code":"4370",
                            "course_name":"ADVANCED TOPICS IN INFORMATION SYSTEMS",
                            "sections": [
                                {
                                    "section_code":"18052",
                                    "unique_id": "mx0ljjklab57"
                                }
                            ]
                        }
                    ]
                }
            ]
        }
    ]
} 
]

我正在使用的 PHP：

$query = "SELECT school_name, term_name, department_name, department_code, course_code, course_name, section_code, magento_course_id
    FROM schools INNER JOIN term_names ON schools.id=term_names.school_id INNER JOIN departments ON schools.id=departments.school_id INNER JOIN adoptions ON departments.id=adoptions.department_id";

$fetch = mysqli_query($con, $query) or die(mysqli_error($con));
$row_array = array();
while ($row = mysqli_fetch_assoc($fetch)) {
  $row_array[$row['school_name']]['school_name'] = $row['school_name'];
  $row_array[$row['school_name']]['terms']['term_name'] = $row['term_name'];
  $row_array[$row['school_name']]['terms']['departments'][] = array(
    'department_name' => $row['department_name'],
    'department_code' => $row['department_code'],
    'course_name' => $row['course_name'],
    'course_code' => $row['course_code'],
    'section_code' => $row['section_code'],
    'unique_id' => $row['magento_course_id']
  );
}

$return_arr = array();
foreach ($row_array as $key => $record) {
  $return_arr[] = $record;
}

file_put_contents("data/iMadeJSON.json" , json_encode($return_arr, JSON_PRETTY_PRINT));

我的 JSON 如下所示：

[
{
    "school_name": "School's Name",
    "terms": {
        "term_name": "FALL 2013",
        "departments": [
            {
                "department_name": "ACCOUNTING",
                "department_code": "ACCT",
                "course_name": "COST ACCOUNTING",
                "course_code": "3315",
                "section_code": "10258",
                "unique_id": "10311"
            },
            {
                "department_name": "ACCOUNTING",
                "department_code": "ACCT",
                "course_name": "ACCOUNTING INFORMATION SYSTEMS",
                "course_code": "3320",
                "section_code": "10277",
                "unique_id": "10314"
            },
            ...

每门课程都重复部门信息，使文件变得更大。我希望更好地了解 PHP 多维数组与 JSON 结合的工作原理，因为我显然不知道。

【问题讨论】：

    标签： php mysql arrays json multidimensional-array


    【解决方案1】：

    将您的 while 更改为：

    while ($row = mysqli_fetch_assoc($fetch)) {
    $row_array[$row['school_name']]['school_name'] = $row['school_name'];
    $row_array[$row['school_name']]['terms']['term_name'] = $row['term_name'];
    $row_array[$row['school_name']]['terms']['department_name'][] = array(
        'department_name' => $row['department_name'],
        'department_code' => $row['department_code']
    );
}

    编辑

    如果你想达到例子中的效果，也许你应该考虑使用这种方法：

    <?php

$result_array = array();

$fetch_school = mysqli_query($con, "SELECT id, school_name FROM schools") or die(mysqli_error($con));
while ($row_school = mysqli_fetch_assoc($fetch_school)) {
    $result_array['school_name'] = $row_school['school_name'];

    $fetch_term = mysqli_query($con, "SELECT term_name FROM term_names WHERE school_id = $row_school['id']") or die(mysqli_error($con));
    while ($row_term = mysqli_fetch_assoc($fetch_term)) {
        $result_array['terms']['term_name'] = $row_term['term_name'];

        $fetch_dept = mysqli_query($con, "SELECT id, department_name, department_code FROM departments WHERE school_id = $row_school['id']") or die(mysqli_error($con));
        while ($row_dept = mysqli_fetch_assoc($fetch_dept)) {
            $result_array['terms']['deptartments']['department_name'] = $row_dept['department_name'];
            $result_array['terms']['deptartments']['department_code'] = $row_dept['department_code'];

            $fetch_course = mysqli_query($con, "SELECT course_name, course_code FROM adoptions WHERE departement_id = $row_dept['id']") or die(mysqli_error($con));
            while ($row_course = mysqli_fetch_assoc($fetch_course)) {
                $result_array['terms']['deptartments']['courses']['course_name'] = $row_course['course_name'];
                $result_array['terms']['deptartments']['courses']['course_code'] = $row_course['course_code'];
            }
        }
    }
}

file_put_contents("data/iMadeJSON.json" , json_encode($result_array, JSON_PRETTY_PRINT));

    可能它不是一个有效的程序，但它应该会给你最好的结果。希望对你有帮助:)

    【讨论】：

    • 这不是我所需要的。您的代码导致每个课程的部门名称重复，但我希望课程嵌套在各自的部门内。 e：不是在，而是在下面，就像我问题开头的 JSON 示例一样。
    • 然后我们需要修改你的整个SELECT查询，先选择父数组，然后加载子数组。我会尽快修改我的答案。
    • @user3798310 我添加了另一种方法。也许你想试一试:)
    • 这绝对是正确的格式。唯一的问题是每个循环都会擦除前面的数组，所以只有最后一个部门的最后一门课程被打印到 JSON 中。看来我比以前更接近了，所以谢谢。
    【解决方案2】：

    尝试用下面的代码替换你的while循环：

    $departments = array();
$courses = array();

$i = 0;
$j = 0;

while ($row = mysqli_fetch_assoc($fetch)) {
    $row_array[$row['school_name']]['school_name'] = $row['school_name'];
    $row_array[$row['school_name']]['terms']['term_name'] = $row['term_name'];

    $key = array_search($row['department_code'], $departments); 
    if ($key === FALSE) {        
        $k = $i++;
        $departments[] = $row['department_code'];
        $row_array[$row['school_name']]['terms']['departments'][$k]['department_name'] = $row['department_name'];
        $row_array[$row['school_name']]['terms']['departments'][$k]['department_code'] = $row['department_code'];
    } else {
        $k = $key;
    }   

    $skey = array_search($row['course_code'], $courses); 
    if ($skey === FALSE) {        
        $l = $j++;
        $courses[] = $row['course_code'];
        $row_array[$row['school_name']]['terms']['departments'][$k]['courses'][$l]['course_name'] = $row['course_name'];
        $row_array[$row['school_name']]['terms']['departments'][$k]['courses'][$l]['course_code'] = $row['course_code'];
    } else {
        $l = $skey;
    } 

    $row_array[$row['school_name']]['terms']['departments'][$k]['courses'][$l]['sections'][] = array('section_code' => $row['section_code'], 'unique_id' => $row['magento_course_id']);
}

    希望这会对你有所帮助。

    【讨论】：

    • 第一个部门完美无缺，但之后的每门课程看起来都是这样的...... "department_name": "ART", "department_code": "ART", "courses": { "28 ": { "course_name": ... "28" 键是问题所在。
    • 能否分享您的表结构和示例数据？
    • 首先，抱歉花了整个周末。其次，我不确定这样做的首选方式是什么，所以这里有一个链接docs.google.com/spreadsheets/d/…
    【解决方案3】：

    我知道这是一个老问题，但今天我遇到了同样的问题。网上没找到合适的解决办法，终于解决了，所以发在这里让别人看看。

    我不能 100% 确定这会奏效，因为我没有您的数据库，但在我的情况下是类似的并且有效。此外，它不会像被要求的那样 100%，但我很确定不会有冗余，所有数据都会显示出来。 

    while ($row = mysqli_fetch_assoc($fetch)) {
$row_array ['school_name'][$row['school_name']]['terms'][$row['term_name']]['departments']['department_code'][$row['department_code']]['department_name'] = $row['department_name'];
$row_array ['school_name'][$row['school_name']]['terms'][$row['term_name']]['departments']['department_code'][$row['department_code']]['courses']['course_code'][$row['course_code']]['course_name'] = $row['course_name'];
$row_array ['school_name'][$row['school_name']]['terms'][$row['term_name']]['departments']['department_code'][$row['department_code']]['courses']['course_code'][$row['course_code']]['sections']['unique_id'][$row['magento_course_id']]['section_code'] = $row['section_code'];
}

    另外，我不是 PHP 人，但据我了解，= 出现在叶子之前且仅在叶子之前。

    【讨论】：

      【解决方案4】：

      我从Ian Mustafa reply开始，想办法解决每个循环擦除前一个数组的问题。

      这是一个旧线程，但我认为这可能对其他人有用，所以这是我的解决方案，但基于我自己的数据结构（很容易弄清楚如何使其适应我认为的其他结构）：

      $usersList_array =array();
$user_array = array();
$note_array = array();

$fetch_users = mysqli_query($mysqli, "SELECT ID, Surname, Name FROM tb_Users WHERE Name LIKE 'G%' ORDER BY ID") or die(mysqli_error($mysqli));
while ($row_users = mysqli_fetch_assoc($fetch_users)) {
    $user_array['id'] = $row_users['ID'];
    $user_array['surnameName'] = $row_users['Surname'].' '.$row_users['Name'];
    $user_array['notes'] = array();

    $fetch_notes = mysqli_query($mysqli, "SELECT id, dateIns, type, content FROM tb_Notes WHERE fk_RefTable = 'tb_Users' AND fk_RefID = ".$row_users['ID']."") or die(mysqli_error($mysqli));
    while ($row_notes = mysqli_fetch_assoc($fetch_notes)) {
        $note_array['id']=$row_notes['id'];
        $note_array['dateIns']=$row_notes['dateIns'];
        $note_array['type']=$row_notes['type'];
        $note_array['content']=$row_notes['content'];
        array_push($user_array['notes'],$note_array);
    }

    array_push($usersList_array,$user_array);
}

$jsonData = json_encode($usersList_array, JSON_PRETTY_PRINT);


echo $jsonData;

      生成的 JSON：

      [
{
    "id": "1",
    "surnameName": "Xyz Giorgio",
    "notes": [
        {
            "id": "1",
            "dateIns": "2016-05-01 03:10:45",
            "type": "warning",
            "content": "warning test"
        },
        {
            "id": "2",
            "dateIns": "2016-05-18 20:51:32",
            "type": "error",
            "content": "error test"
        },
        {
            "id": "3",
            "dateIns": "2016-05-18 20:53:00",
            "type": "info",
            "content": "info test"
        }
    ]
},
{
    "id": "2",
    "cognomeNome": "Xyz Georg",
    "notes": [
        {
            "id": "4",
            "dateIns": "2016-05-20 14:38:20",
            "type": "warning",
            "content": "georg warning"
        },
        {
            "id": "5",
            "dateIns": "2016-05-20 14:38:20",
            "type": "info",
            "content": "georg info"
        }
    ]
}
]

      【讨论】：

      • 像宝石一样工作，为我节省了大量时间，谢谢！
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