如何使用PHP编码多个json元素？

Question

我试图在Android中创建一个问卷调查应用程序，我已经可以从数据库中获取问题，并将其显示为JSON，但它将以以下格式显示：

​

{
    "result": [
        {
            "id_question": "1",
            "question_name": "Grade level",
            "choices": "Grade 11, Grade 12"
        },
        {
            "id_question": "2",
            "question_name": "Expected grade in this subject",
            "choices": "90-100, 75-89, 60-74, Below 60"
        }
    ]
}

​

但是，我在Android中使用的库只接受这种JSON格式来显示问题：

​

{
  "survey_properties": {
    "intro_message": "To get a reliable result for the evaluation, please respond to all questions.",
    "end_message": "Your answers have been recorded. <br>Thank you for taking the time to answer the evaluation."
  },
  "questions": [
    {
      "id_question": "1",
      "question_name": "Grade Level",
      "choices": [
        "Grade 11",
        "Grade 12"
      ]
    },
    {
      "id_question": "2",
      "question_name": "Expected Grade in this subject",
      "choices": [
        "90-100",
        "75-89",
        "60-74",
        "Below 60"
      ]
    }
  ]
}

​

如何在PHP中实现这种输出？这是我正在使用的脚本：

​

$query = "SELECT * FROM question_test";

$r = mysqli_query($conn, $query);
$result = array();

while($row = mysqli_fetch_array($r)) {
    array_push($result,array(
        "id_question"=>$row['id_question'],
        "question_name"=>$row['question_name'],
        "choices"=>$row['choices']
        )
    );
}

echo json_encode(array("result"=>$result));

​

Answer 1

从您的代码中，可以这样修改：

$query = "SELECT * FROM question_test";

$r = mysqli_query($conn, $query);
$result = array();

while ($row = mysqli_fetch_array($r)) {
    array_push($result,array(
        "id_question"=>$row['id_question'],
        "question_name"=>$row['question_name'],
        "choices"=>explode(', ', $row['choices'])
    ));
}

echo json_encode(array("result"=>$result));

您必须添加更多像survey_properties这样的信息

Answer 2

我在这里要做的是把“选择”放到一个数组中。希望能成功。

$query = "SELECT * FROM question_test";

$r = mysqli_query($conn, $query);
$result = array();
$choices = array();

while($row = mysqli_fetch_array($r)) {
    array_push($result,array(
        "id_question"=>$row['id_question'],
        "question_name"=>$row['question_name'],
        array_push($choices,array($row['choices'])
        )
    );
}

echo json_encode(array("result"=>$result));