php 使用php读取Json数据响应
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Read Json data response using php
提问by rocksphp
How can I read a JSON data response using php? The response t comes after user authentication done from a third party. Primarily, I just want displayNameand preferredUsernamedata.
如何使用 php 读取 JSON 数据响应？响应 t 在第三方完成用户身份验证后出现。首先，我只想要displayName和preferredUsername数据。
Json response:
杰森回复：
{
"stat": "ok",
"profile": {
"providerName": "testing",
"identifier": "http://testing.com/58263223",
"displayName": "testing",
"preferredUsername": "testing",
"name": {
"formatted": "testing"
},
"url": "http://testing.com/testing/",
"photo": "https://securecdn.testing.com/uploads/users/5826/3223/avatar32.jpg?1373393837",
"providerSpecifier": "testing"
}
}
回答by vee
You can use the json_decode(http://php.net/manual/en/function.json-decode.php) function to decode your result then retrieve the value:
您可以使用json_decode( http://php.net/manual/en/function.json-decode.php) 函数解码您的结果然后检索值：
$json_data = '{
"stat": "ok",
"profile": {
"providerName": "testing",
"identifier": "http://testing.com/58263223",
"displayName": "testing",
"preferredUsername": "testing",
"name": {
"formatted": "testing"
},
"url": "http://testing.com/testing/",
"photo": "https://securecdn.testing.com/uploads/users/5826/3223/avatar32.jpg?1373393837",
"providerSpecifier": "testing"
}
}';
$json = json_decode($json_data);
echo $json->profile->displayName;
echo $json->profile->preferredUsername;
回答by vee
<?php
$json='{
"stat": "ok",
"profile": {
"providerName": "testing",
"identifier": "http://testing.com/58263223",
"displayName": "testing",
"preferredUsername": "testing",
"name": {
"formatted": "testing"
},
"url": "http://testing.com/testing/",
"photo": "https://securecdn.testing.com/uploads/users/5826/3223/avatar32.jpg?1373393837",
"providerSpecifier": "testing"
}
}';
$data=json_decode($json ,true);
$preferredUsername=$data['profile']['preferredUsername'];
$displayName=$data['profile']['displayName'];
?>
回答by eclipsis
json_decodeis what you're looking for:
json_decode是您要查找的内容：
$json = '[
{
"displayName": "testing",
"preferredUsername": "testing",
}
]';
$jsonArray = json_decode($json);
foreach($jsonArray as $value){
$displayName = $value->Display Name;
$preferredUsername = $value->Preferred User;
}
