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我已经编写了一个脚本，除了文件的实际上传之外，一切似乎都很好。这是我拥有的php代码：

<?php
$allowedExts = array("gif", "jpeg", "jpg", "png");
$extension = end(explode(".", $_FILES["avatar"]["name"]));
if ((($_FILES["avatar"]["type"] == "image/gif")
|| ($_FILES["avatar"]["type"] == "image/jpeg")
|| ($_FILES["avatar"]["type"] == "image/jpg")
|| ($_FILES["avatar"]["type"] == "image/pjpeg")
|| ($_FILES["avatar"]["type"] == "image/x-png")
|| ($_FILES["avatar"]["type"] == "image/png"))
&& ($_FILES["avatar"]["size"] < 20000)
&& in_array($extension, $allowedExts))
{
    if ($_FILES["avatar"]["error"] > 0)
    {
    header("Location: ../../dashboard.php?page=profile&error=true");
    }
    else
    {
    $newfilename = "uwhduwhd";
    $path = "../../avatar/";

    move_uploaded_file($_FILES["avatar"]["tmp_name"],"$path" . $newfilename); 

    header("Location: ../../dashboard.php?page=profile&error=false");
    }
}
else
{
header("Location: ../../dashboard.php?page=profile&error=true");
}
?>

我还想做的是将 $newfilename 作为通过上一页上的表单传递的 ID，使用

有人可以帮忙吗？谢谢

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