试试这个：

var_export( json_decode( '[{"location":[{"building":["Default Building"],"name":"Default Location"}],"name":"Default Organization"}]' )  );

json_decode返回数组或object .你可以打印它与var_export不是echo
您可以访问值：

$items = json_decode('[{"location":[{"building":["Default Building"],"name":"Default Location"}],"name":"Default Organization"}]');

foreach( $items as $each ){
  echo $each->location[0]->building[0];
  echo '<hr />';
  echo $each->location[0]->name;
  echo '<hr />';
  echo $each->name; // default organization
}

你的json是有效的，可能是你在访问数组中的对象时遇到了问题。
print_r是理解数组结构的好朋友。试试这个

$json = '[{"location":[{"building":["Default Building"],"name":"Default Location"}],"name":"Default Organization"}]';
$decoded = json_decode($json);

echo '<pre>';
print_r($decoded);

$location = $decoded[0]->location;
$building = $location[0]->building[0];
$name = $location[0]->name;

对象将只返回第一项，如果数组有多个值，则使用foreach

这是一个有效的JSON。

$my_json = '[{"location":[{"building":["Default Building"],"name":"Default Location"}],"name":"Default Organization"}]';
$my_data = json_decode($my_json);
print_r($my_data);

//输出

Array
(
    [0] => stdClass Object
        (
            [location] => Array
                (
                    [0] => stdClass Object
                        (
                            [building] => Array
                                (
                                    [0] => Default Building
                                )

                            [name] => Default Location
                        )

                )

            [name] => Default Organization
        )

)

在这种情况下，我更喜欢在字符串前后添加花括号。它允许我使用json_decode($json, true);，这是我最喜欢的与json变量交互的方式。

$my_json = '[{"location":[{"building":["Default Building"],"name":"Default Location"}],"name":"Default Organization"}]';
  $json = json_decode('{ "data":'.$my_json.'}', true);
  $my_data = $json['data'][0];
  print_r($my_data['location']);
  echo $my_data['location'][0]['building'][0];