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我有一个文件用作PHP来充当配置文件以存储可能需要频繁更改的信息。我返回数组作为对象,像这样:尝试从返回的对象获取非对象的属性
return (object) array(
"host" => array(
"URL" => "https://thomas-smyth.co.uk"
),
"dbconfig" => array(
"DBHost" => "localhost",
"DBPort" => "3306",
"DBUser" => "thomassm_sqlogin",
"DBPassword" => "SQLLoginPassword1234",
"DBName" => "thomassm_CadetPortal"
),
"reCaptcha" => array(
"reCaptchaURL" => "https://www.google.com/recaptcha/api/siteverify",
"reCaptchaSecretKey" => "IWouldNotBeSecretIfIPostedItHere"
)
);
在我的班级我有一个构造函数调用此: 私人$配置;
function __construct(){
$this->config = require('core.config.php');
}
而且使用它像:
curl_setopt($ch, CURLOPT_POSTFIELDS, http_build_query(array('secret' => $this->config->reCaptcha->reCaptchaSecretKey, 'response' => $StrToken)));
不过,我给出的错误:
[18-Apr-2017 21:18:02 UTC] PHP Notice: Trying to get property of non-object in /home/thomassm/public_html/php/lib/CoreFunctions.php on line 21
我不明白为什么会这样考虑的事情就是返回一个对象,它似乎适用于其他人,因为我从另一个问题得到了这个想法。有什么建议么?
有什么stdClass的东西? –
这就是对象。无论何时通过强制转型创建对象,或者通过未指定类的源创建对象,它都属于'stdClass'类。 – AbraCadaver