如何在PHP中打印JSON数据

我想从下面的JSON结果中打印这些键“name”、“adminName1”、“countryCode”

{
"geonames": [{
    "adminCode1": "16",
    "lng": "74.19774",
    "distance": "2.95838",
    "geonameId": 1254611,
    "toponymName": "Thengode",
    "countryId": "1269750",
    "fcl": "P",
    "population": 0,
    "countryCode": "IN",
    "name": "Thengode",
    "fclName": "city, village,...",
    "countryName": "India",
    "fcodeName": "populated place",
    "adminName1": "Maharashtra",
    "lat": "20.51997",
    "fcode": "PPL"
}]
}

---

如何正确地执行它？

只需使用

json\u解码

并获取值即可。由于

json\u decode

函数返回一个对象数组，因此需要使用

->

来访问值

如果您的json字符串名称是

$json

$arr = json_decode($json);

echo $arr->geonames[0]->name; //Thengode
echo $arr->geonames[0]->adminName1; //Maharashtra
echo $arr->geonames[0]->countryCode; //IN

您的json解码数组将是：

stdClass Object
(
    [geonames] => Array
        (
            [0] => stdClass Object
                (
                    [adminCode1] => 16
                    [lng] => 74.19774
                    [distance] => 2.95838
                    [geonameId] => 1254611
                    [toponymName] => Thengode
                    [countryId] => 1269750
                    [fcl] => P
                    [population] => 0
                    [countryCode] => IN
                    [name] => Thengode
                    [fclName] => city, village,...
                    [countryName] => India
                    [fcodeName] => populated place
                    [adminName1] => Maharashtra
                    [lat] => 20.51997
                    [fcode] => PPL
                )

        )

)

---

这很简单，看看吧

<?php 

//just for error loggin
error_reporting(1);
ini_set('display_errors', true);

//json data
$json = '{
"geonames": [{
    "adminCode1": "16",
    "lng": "74.19774",
    "distance": "2.95838",
    "geonameId": 1254611,
    "toponymName": "Thengode",
    "countryId": "1269750",
    "fcl": "P",
    "population": 0,
    "countryCode": "IN",
    "name": "Thengode",
    "fclName": "city, village,...",
    "countryName": "India",
    "fcodeName": "populated place",
    "adminName1": "Maharashtra",
    "lat": "20.51997",
    "fcode": "PPL"
}]}';

//decoded json object
$jsonDataObject = json_decode($json);

//parsed variables
$name = $jsonDataObject->geonames[0]->name;
$adminName1 = $jsonDataObject->geonames[0]->adminName1;
$countryCode = $jsonDataObject->geonames[0]->countryCode;
?>

在尝试了你的代码之后，我收到了来自PHP的这封情书：-D PHP注意：尝试在线获取非objectcheck的属性：谢谢Frayne和Dan:-）尝试添加所有必需的字段…将JSON解码到数组中，使用

array_keys（）

获取密钥，然后获取你需要的密钥，先生，如果他需要这些钥匙，那么为什么需要解码？？他早就认识他们了。

<?php 

//just for error loggin
error_reporting(1);
ini_set('display_errors', true);

//json data
$json = '{
"geonames": [{
    "adminCode1": "16",
    "lng": "74.19774",
    "distance": "2.95838",
    "geonameId": 1254611,
    "toponymName": "Thengode",
    "countryId": "1269750",
    "fcl": "P",
    "population": 0,
    "countryCode": "IN",
    "name": "Thengode",
    "fclName": "city, village,...",
    "countryName": "India",
    "fcodeName": "populated place",
    "adminName1": "Maharashtra",
    "lat": "20.51997",
    "fcode": "PPL"
}]}';

//decoded json object
$jsonDataObject = json_decode($json);

//parsed variables
$name = $jsonDataObject->geonames[0]->name;
$adminName1 = $jsonDataObject->geonames[0]->adminName1;
$countryCode = $jsonDataObject->geonames[0]->countryCode;
?>