如何自定义 PHP 中的 json_encode 数据？

Question

我有一个名为questions的 MySQL 数据库表，具有以下列：

id, question, category, type, difficulty, correct_answer, incorrect_answer_1, incorrect_answer_2, incorrect_answer_3

我需要将此表中的 output 数据转换为JSON格式，但仅采用以下特定格式（如错误答案合并在一起）：

{
  "response_code": 0,
  "results": [
    {
      "category": "Science & Nature",
      "type": "multiple",
      "difficulty": "hard",
      "question": "An organic compound is considered an alcohol if it has what functional group?",
      "correct_answer": "Hydroxyl",
      "incorrect_answers": [
        "Carbonyl",
        "Alkyl",
        "Aldehyde"
      ]
    },
    {
      "category": "Entertainment: Video Games",
      "type": "multiple",
      "difficulty": "easy",
      "question": "Which of the following is not a faction in Tom Clancy's The Division?",
      "correct_answer": "CDC",
      "incorrect_answers": [
        "Cleaners",
        "Last Man Batallion",
        "Rikers"
      ]
    },
    {
      "category": "Entertainment: Video Games",
      "type": "multiple",
      "difficulty": "medium",
      "question": "Which of the following Call of Duty games was a PS3 launch title?",
      "correct_answer": "Call of Duty 3",
      "incorrect_answers": [
        "Call of Duty 4: Modern Warfare",
        "Call of Duty: World at War",
        "Call of Duty: Roads to Victory"
      ]
    }
  ]
}

With the basic knowledge of PHP and json_encode , I can output the data in JSON format, but unable to output it in the above-mentioned format. 我的 PHP 代码是：

$mysqli = new mysqli("localhost", "root", "", "demo_db");
$statement = $mysqli->prepare("SELECT * FROM questions limit 50");
$statement->execute(); 
$result = $statement->get_result();
$outp = $result->fetch_all(MYSQLI_ASSOC);
echo json_encode($outp); // Tried this, but it's not outputting in desired format

//echo json_encode(($result->fetch_assoc())); // Tried this, but it's also not outputting in desired format
//echo json_encode(($result->fetch_array())); // Tried this, but it's also not outputting in desired format

我尝试了三种不同的方式来使用json_encode ，但我无法制作所需的格式，特别是使不正确的答案成为类似嵌套数组的格式。

我应该怎么办？

Answer 1

您在想要的 output 中有字段，结果中没有返回，因此您必须自己构建。

这只是一个指导性的答案！ 这不会产生完整的输出！ 您必须自己处理详细信息，尤其是对于incorrect_answers的答案。

$results = [
    'response_code' => 0,
    'results' => []
];

foreach($outp as $item) {
    $results['results'][] = $item;
}

echo json_encode($results, JSON_PRETTY_PRINT);

Answer 2

基本上，您希望一次检索一个结果集中的行（例如，在while循环中），然后将检索到的行转换为您想要的格式并将其添加到output变量中。

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$mysqli = new mysqli("localhost", "root", "", "demo_db");

$sql = "
    SELECT
        id,
        question,
        category,
        type,
        difficulty,
        correct_answer,
        incorrect_answer_1,
        incorrect_answer_2,
        incorrect_answer_3
    FROM questions
    Limit 50
";

$query  = $mysqli->prepare($sql);
$query->execute();
$result = $query->get_result();

$output = [];


// Loop through the result set row by row
while ($row = $result->fetch_assoc()) {
    $output[] = [
        "category"          => $row["category"],
        "type"              => $row["type"],
        "difficulty"        => $row["difficulty"],
        "question"          => $row["question"],
        "correct_answer"    => $row["correct_answer"],
        "incorrect_answers" => [
            $row["incorrect_answer_1"],
            $row["incorrect_answer_2"],
            $row["incorrect_answer_3"],
        ],
    ];
}

// Convert to JSON and print
echo json_encode($output);