PHP convet class to json data

標籤：tac   輸出   obj   自動   port   內容   usr   https   context   

/********************************************************************* *              PHP convet class to json data * 說明： *     突然想使用class自動轉換為json資料，這樣的代碼可擴充性會好一點， * 只需要修改class的屬性就能夠達到最終json資料輸出，不過有遇到class中 * 初始化class變數需要在建構函式中初始化的的問題。 * *                                   2017-8-11 深圳 龍華樟坑村 曾劍鋒 ********************************************************************/一、參考文檔：    1. getting Parse error: syntax error, unexpected T_NEW [closed]        https://stackoverflow.com/questions/15806981/getting-parse-error-syntax-error-unexpected-t-new二、測試代碼：    <?php        class Uart {            public $port = "/dev/ttyO0";            public $value = "OK";        }        class Context {            public $uart = new Uart();;            public $version = "v0.0.1";        }        $context = new Context;        $context_json = json_encode($context);        echo $context_json    ?>三、報錯內容：    Parse error: syntax error, unexpected ‘new‘ (T_NEW) in /usr/share/web/time.php on line 8四、最終代碼：    <?php        class Uart {            public $port = "/dev/ttyO0";            public $value = "OK";        }        class Context {            public $uart;            public $version = "v0.0.1";            public function __construct() {                $this->uart = new Uart();            }        }        $context = new Context;        $context_json = json_encode($context);        echo $context_json    ?>五、輸出結果：    {"uart":{"port":"\/dev\/ttyO0","value":"OK"},"version":"v0.0.1"}六、原因：    you must do initialize new objects in the __construct function;

 

PHP convet class to json data