如何访问在 php 中命名为变量的对象属性?
以 JSON 编码的 Google API 返回了这样的对象
A Google APIs encoded in JSON returned an object such as this
[updated] => stdClass Object
(
[$t] => 2010-08-18T19:17:42.026Z
)
谁知道我如何访问 $t 值?
Anyone knows how can I access the $t value?
$object->$t 明显返回
注意:未定义变量:t in/usr/local/...
Notice: Undefined variable:
tin /usr/local/...
致命错误:无法访问/.... 中的空属性
Fatal error: Cannot access empty property in /....
推荐答案
由于您的属性名称是字符串 '$t'，您可以这样访问它:
Since the name of your property is the string '$t', you can access it like this:
echo $object->{'$t'};
或者，您可以将属性的名称放在一个变量中并像这样使用它:
Alternatively, you can put the name of the property in a variable and use it like this:
$property_name = '$t';
echo $object->$property_name;
您可以在 repl.it 上看到这两种操作:https://repl.it/@jrunning/SpirtedTroubledWorkspace
You can see both of these in action on repl.it: https://repl.it/@jrunning/SpiritedTroubledWorkspace
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