Strct标准：仅变量应通过引用传递

Question

我有这个代码

$maxSize="44400000";
$allowedExts = array("jpg","jpeg","docx","png","JPG");
$extension = end(explode(".", $_FILES["file"]["name"])); 

并给我下面显示的错误。

主要问题是上述行。

严格的标准：只能通过引用传递变量

Answer 1

您必须再创建一个变量：

$maxSize="44400000";
$allowedExts = array("jpg","jpeg","docx","png","JPG");
$extension = explode(".", $_FILES["file"]["name"]);
$final_ext = end($extension);

但是最好使用pathinfo来做到这一点。

$path = $_FILES['file']['name'];
$ext = pathinfo($path, PATHINFO_EXTENSION);

Answer 2

由于最终函数仅接受必须通过引用传递的参数，因此您应该这样做：

$maxSize="44400000";
$allowedExts = array("jpg","jpeg","docx","png","JPG");
$explode = explode(".", $_FILES["file"]["name"]);
$extension = end($explode);

了解更多http://php.net/manual/zh/function.end.php