使用 PHP 提取 JSON 数据

Question

假设下面是我的 JSON 数据

{"pricing": {
     "com": {
         "addons": {
                "dns": true,
                "email": true,
                "idprotect": true
       },
     "org": {
         "addons": {
                "dns": true,
                "email": true,
                "idprotect": true
       },
     "net": {
         "addons": {
                "dns": true,
                "email": true,
                "idprotect": true
       }
}}

我只想从 JSON 上方显示（com、org、net）。 我们怎么做？

Answer 1

这样做的一个选项是使用json_decode并为第二个参数传递true以将返回的对象转换为关联数组。

要仅显示键，您可以使用foreach循环$output["pricing"]并显示键：

$json = '{"pricing": {"com": {"addons": {"dns": true,"email": true,"idprotect": true}},"org": {"addons": {"dns": true,"email": true,"idprotect": true}},"net": {"addons": {"dns": true,"email": true,"idprotect": true}}}}';
$output = json_decode($json, true);

foreach ($output["pricing"] as $key => $value) {
    echo $key . "<br>";
}

另一种方法是获取array_keys并循环它们：

foreach (array_keys($output["pricing"]) as $key) {
    echo $key . "<br>";
}

Answer 2

首先你有一个不正确的 JSON 格式检查出来。 我认为这个解决方案可能对你有帮助！！

$json = '{"pricing": {"com": {"addons": {"dns": true,"email": true,"idprotect": true}},"org": {"addons": {"dns": true,"email": true,"idprotect": true}},"net": {"addons": {"dns": true,"email": true,"idprotect": true}}}}';
$output = json_decode($json);

print_r($output->pricing->com->addons);
print_r($output->pricing->org->addons);
print_r($output->pricing->net->addons);

你会得到这样的东西！！

stdClass Object ( [dns] => 1 [email] => 1 [idprotect] => 1 ) 
stdClass Object ( [dns] => 1 [email] => 1 [idprotect] => 1 ) 
stdClass Object ( [dns] => 1 [email] => 1 [idprotect] => 1 )

Answer 3

你在找这个吗？？

 $json = '{"pricing": {"com": {"addons": {"dns": true,"email": true,"idprotect": true}},"org": {"addons": {"dns": true,"email": true,"idprotect": true}},"net": {"addons": {"dns": true,"email": true,"idprotect": true}}}}';
$output = json_decode($json,true);

echo implode(",",array_keys($output["pricing"]));

com,org,net