在PHP中解析此JSON对象

我将通过HTTPPOST接收一个JSON对象，我发现很难解析它。这就是JSON对象的外观：

{ login: {username: 123, password: 456} }, questions:[{ name: "insomnia", type: "boolean", problem: true, question: "Did you experience    insomnia?", answer: null},{ name: "go-to-bed", type: "amount", problem: false, question: "When did you go to bed?", answer: null }]}

我想把它解析成3个不同的变量$username、$password和$q

从这个例子来看，这就是我所期望的：

echo$username/**输出：*123

echo$password//**输出：*456

---

echo$q/**output:**questions:[{name:“失眠”，键入：“布尔”，问题：true，问题：“你经历过失眠吗？”，回答：null}，{name:“上床睡觉”，键入：“金额”，问题：false，问题：“你什么时候上床了？”，回答：null}]

首先，你的示例不是有效的json。这里是有效的一个：

[{
"login": {
    "username": 123,
    "password": 456
},
"questions": [{
    "name": "insomnia",
    "type": "boolean",
    "problem": true,
    "question": "Did you experience    insomnia?",
    "answer": null
}, {
    "name": "go-to-bed",
    "type": "amount",
    "problem": false,
    "question": "When did you go to bed?",
    "answer": null
}]
}]

接下来，您可以从字符串中使用

json\u decode

：

$x = '[{
"login": {
    "username": 123,
    "password": 456
},
"questions": [{
    "name": "insomnia",
    "type": "boolean",
    "problem": true,
    "question": "Did you experience    insomnia?",
    "answer": null
}, {
    "name": "go-to-bed",
    "type": "amount",
    "problem": false,
    "question": "When did you go to bed?",
    "answer": null
}]
}]';

$q = json_decode($x);
print_r($q);
echo $q[0]->login->username;

---

示例中的JSON无效，我修复了它并进行了测试，我认为这就是您要寻找的

<?php
$json = <<<JSON
{
  "login": {
    "username": 123,
    "password": 456
  },
  "questions": [
    {
      "name": "insomnia",
      "type": "boolean",
      "problem": true,
      "question": "Did you experience insomnia?",
      "answer": null
    },
    {
      "name": "go-to-bed",
      "type": "amount",
      "problem": false,
      "question": "When did you go to bed?",
      "answer": null
    }
  ]
}
JSON;

$decoded = json_decode($json);

$username = $decoded->login->username;
$password = $decoded->login->password;

// Re-encode questions to a JSON string
$q = json_encode($decoded->questions);

echo $username."\n";
echo $password."\n";
echo $q."\n"; 

---

？@JonStirling尝试过，但问题是解析存储在$q中的对象并将其存储为字符串。我不知道如何在不改变输入完整性的情况下将子JSON对象存储为字符串。我对PHP很陌生。这可能真的很容易。你能把它编码吗<代码>$q=json_编码（$decoded['questions']）到目前为止您有代码吗？小提示：这不是有效的JSONSTRING echo语句不打印anything@GabrielAluyor对不起，我忘了加引号了。我编辑了这篇博文，这是为我做的，但我不得不做一些修改。echo$q[0][login][username]；echo$q[0][登录][密码]；echo json_encode（$q[0][questions]）；但不幸的是，我不能推翻这个答案——声誉不够。@GabrielAluyor我想你是想回答另一个答案的。