php - 使用 json_encode 从 MySQL 查询返回 JSON 对象

Question

问题很长，所以我不得不缩短它。

无论如何，我目前有下表，结果如下。

我正在做的是:

1. 查询与一个问题相关的所有答案
2. 存入数组后编码

这是我当前的查询:

$stmt = "SELECT questions.question_text, answers.answer_text 
   FROM     questions, answers, test
   WHERE    questions.question_id = answers.question_id
   AND      questions.test_id =1";

$result = $connection->query($stmt);

这给了我这个:

这是 PHP:

$encode = array();

while($row = mysqli_fetch_assoc($result)) {
   $encode[] = $row;
}

echo json_encode($encode);  

这给了我这个输出:

[
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is a Hypertext Markup Language"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is a Hypertext Markup Language"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is a food"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is a food"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is an Asynchronous language"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is an Asynchronous language"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is a styling language"
    },
    {
        "question_text": "What is HTML?",
        "answer_text": "HTML is a styling language"
    }
]

这是使用 json_encode 的所需输出:

"What is HTML?": {
        "1": "HTML is a Hypertext Markup Language",
        "2": "HTML is a food",
        "3": "HTML is an Asynchronous language",
        "4": "HTML is a styling language"
    }

我目前得到的是多个单个 对象，其中包含一个答案，但始终是与之关联的答案。我希望制作一个包含所有答案的对象以及代表该对象的问题。我真的希望这是有道理的。我的逻辑可能有偏差，所以请原谅我。

我尝试使用 while 循环，但无法让它工作。有人可以引导我以正确的方式实现我想要的输出吗？

谢谢。

Answer 1

听起来就像只是改变你正在构建的数组......

$encode = array();

while($row = mysqli_fetch_assoc($result)) {
   $encode[$row['question _text']][] = $row['answer_text'];
}

echo json_encode($encode);