Php 有没有办法像这样从MySQL解析JSON

我想从服务器创建一个PHP API，将JSON解析为Android 但没关系

主要的问题是我如何编写代码

这是我的PHP代码

<?php

include 'includes/conn.php';


$sql = "SELECT * FROM positions ORDER BY priority ASC";

$res = array();

        $query = $conn->query($sql);
        while($row = $query->fetch_assoc()){

            $csql = "SELECT * FROM candidates WHERE position_id='".$row['id']."'";
            $cquery = $conn->query($csql);

            $cs = array( 
              'pos_id' => $row['id'],
                'pos_des' => $row['description'],
                'max_votes' => $row['max_vote']);

            while($crow = $cquery->fetch_assoc()){

                    $fe = array(
                      'candidates' => array(
                     'id' => $crow['id'],
                     'pos_id' => $crow['position_id'],
                     'firstname' => $crow['firstname'],
                     'lastname' => $crow['lastname'],
                     'photo' => $crow['photo'],
                     'platform' => $crow['platform']

                     array_push($cs,fe);

                 )          
            );

        }

        array_push($res,$cs);

}

    echo json_encode($res);

?>

我想要这个结果

  [{
        "pos_id": "1",
        "pos_des": "Reprisentative",
        "max_votes": "4",
        "candidates": [{
            "id": "1",
            "pos_id": "1",
            "firstname": "G-9-JAY",
            "lastname": "MCARTHUR",
            "photo": "",
            "platform": ""
        }, {
            "id": "2",
            "pos_id": "1",
            "firstname": "G-10-JOHN",
            "lastname": "MARTIN",
            "photo": "",
            "platform": ""
        }, {
            "id": "5",
            "pos_id": "1",
            "firstname": "GRADE 7",
            "lastname": "Lang ya",
            "photo": "",
            "platform": ""
        }]

    },
    {
        "pos_id": "2",
        "pos_des": "President",
        "max_votes": "1",
        "candidates": [{
            "id": "3",
            "pos_id": "2",
            "firstname": "User",
            "lastname": "Name",
            "photo": "",
            "platform": ""
        }, {
            "id": "4",
            "pos_id": "2",
            "firstname": "TEST",
            "lastname": "USER",
            "photo": "",
            "platform": ""
        }]
    }
  ]

小结：我希望用户输入他/她分配给职位的职位id，例如，MARTIN是一位竞选总统，因此MARTIN应该显示在总统职位上

---

如何实现这一点？

您应该使用

$cs['candidates']

，而不是创建一个单独的数组，将

候选者

作为每个候选者的键

            $cs = array( 
                'pos_id' => $row['id'],
                'pos_des' => $row['description'],
                'max_votes' => $row['max_vote'],
                'candidates' => array());

            while($crow = $cquery->fetch_assoc()){
                $cs['candidates'][] = array(
                    'id' => $crow['id'],
                    'pos_id' => $crow['position_id'],
                    'firstname' => $crow['firstname'],
                    'lastname' => $crow['lastname'],
                    'photo' => $crow['photo'],
                    'platform' => $crow['platform']
                );
            }

---

您应该使用

$cs['candidates']

，而不是创建一个单独的数组，其中

candidates

作为每个候选项的键

            $cs = array( 
                'pos_id' => $row['id'],
                'pos_des' => $row['description'],
                'max_votes' => $row['max_vote'],
                'candidates' => array());

            while($crow = $cquery->fetch_assoc()){
                $cs['candidates'][] = array(
                    'id' => $crow['id'],
                    'pos_id' => $crow['position_id'],
                    'firstname' => $crow['firstname'],
                    'lastname' => $crow['lastname'],
                    'photo' => $crow['photo'],
                    'platform' => $crow['platform']
                );
            }

---

. 错误：语法错误，第23I行中意外的“候选者”（T_常量_封装的_字符串），应为“）”缺少逗号。。错误：语法错误，第23I行中意外的“候选者”（T_常量_封装的_字符串），应为“）”缺少逗号。