Question

这是我使用CURL从API中捕获的字符串：

{
  "ip": "8.8.8.8",
  "hostname": "google-public-dns-a.google.com",
  "city": "Mountain View",
  "region": "California",
  "country": "US",
  "loc": "37.3860,-122.0838",
  "org": "AS15169 Google Inc.",
  "postal": "94040"
}

如何获取国家/地区代码？这就是我一直在尝试的，它只返回整个json字符串：

$json = json_decode($json_raw, true);
    $country = $json['country'];

Answer 1

由于您在对象中输入值，请使用->进行访问。

<?php
$a = '{
  "ip": "8.8.8.8",
  "hostname": "google-public-dns-a.google.com",
  "city": "Mountain View",
  "region": "California",
  "country": "US",
  "loc": "37.3860,-122.0838",
  "org": "AS15169 Google Inc.",
  "postal": "94040"
}';
$decode = json_decode($a);

echo $decode->country;// this line

请参阅演示here

正如Surace所说，即使这应该可以正常工作

<?php
    $a = '{
       "ip": "8.8.8.8",
       "hostname": "google-public-dns-a.google.com",
       "city": "Mountain View",
       "region": "California",
       "country": "US",
       "loc": "37.3860,-122.0838",
       "org": "AS15169 Google Inc.",
       "postal": "94040"
  }';
$decode = json_decode($a,true);

echo $decode['country'];

请参阅演示here

Answer 2

当您将json解码为php格式时，它会将结果作为对象返回，因此您必须使用对象syn文本来访问其属性。替换

  $decode = json_decode($a,true);
    $country = $json['country'];

<强>与

 $decode = json_decode($a);
     $country = $json->country;

Answer 3

我可能会遗漏一些东西，但我认为你的代码没有理由不适合你。您可以使用json_decode和true选项一样使用返回值作为数组。

<?php
$json_raw = '{
"ip": "8.8.8.8",
"hostname": "google-public-dns-a.google.com",
"city": "Mountain View",
"region": "California",
"country": "US",
"loc": "37.3860,-122.0838",
"org": "AS15169 Google Inc.",
"postal": "94040"
}';

$json = json_decode($json_raw, true);
echo $country = $json['country'];

您可以在此处查看：https://eval.in/483744

如何从此JSON字符串获取国家/地区？

3 个答案: