[英]Parse this JSON OBJECT in PHP
我將通過HTTP POST接收JSON對象,並且發現很難解析它。 JSON對象如下所示:
{ 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 }]}
我想將其解析為3個不同的變量$ username,$ password和$ q
從示例中,這就是我期望的結果:
echo $username // **output:** 123
echo $password // **output:** 456
echo $q //**output:** 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 }]
首先,您的示例不是有效的json。 這里是有效的:
[{
"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
}]
}]
接下來,您可以從字符串中使用json_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;
該示例中的JSON無效,我對其進行了修復並進行了測試,我認為這就是您想要的。
<?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";
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