PHP：从对象中获取数据

Question

如何从该对象获取 user_nicename ？

BP_User Object
(
    [data] => stdClass Object
        (
            [ID] => 1
            [user_login] => NICENICE
            [user_pass] => $P$BwLHvV7zxcZZ/zW7MY0NXNSmANP.U5.
            [user_nicename] => NICENAME
            ...

我在哪里可以找到学习此内容的资源？

Answer 1

$variable->data->user_nicename

应该有效。

Answer 2

您也可以通过另一种方式打印

foreach($result->data as $newdata) //$result is variable that store raw data 
{
printf("name: %s, Pass: %s, nicename: %s <br/>", $newdata->user_login, $newdata->user_pass, $newdata->user_nicename);
}

结果将是

name:NICENICE , Pass:$P$BwLHvV7zxcZZ/zW7MY0NXNSmANP.U5., nicename:NICENAME

Answer 3

我有这种类型的数据。我已经像这样从数组中提取了数据。
该类型数据

$original = new DateTime("2023-04-28T14:51:12-07:00");
$data = json_decode(json_encode($original));
$time = strtotime($data->date);
echo $time;