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
       }
}}

我想从JSON上面只显示（com，org，net）。我们怎么能这样做？

Answer 1

执行此操作的选项是使用json_decode并将true传递给第二个参数，以将返回的对象转换为关联数组。

要仅显示您可以使用foreach循环$output["pricing"]的密钥并显示密钥：

$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"]));

玉米，组织，净

Answer 4

您可以轻松完成：

$parsed_json= json_decode("your raw data here");
$com = $parsed_json["com"];
$org = $parsed_json["org"];
$net = $parsed_json["net"];

您可以在PHP文档网站上了解有关json_decode的更多信息。

使用PHP提取JSON数据

4 个答案: