PHP上传文件并给出一个自定义名称

Question

我正在使用此函数上传我的文件：

if ((($_FILES["Artwork"]["type"] == "image/gif")
|| ($_FILES["Artwork"]["type"] == "image/jpeg")
|| ($_FILES["Artwork"]["type"] == "image/jpg")
|| ($_FILES["Artwork"]["type"] == "image/pjpeg"))
&& ($_FILES["Artwork"]["size"] < 20000000))
  {
  if ($_FILES["Artwork"]["error"] > 0)
    {
    //echo "Return Code: " . $_FILES["Artwork"]["error"] . "<br />";
    }else{
      $imageName = $_FILES['Artwork']['name'];
      move_uploaded_file($_FILES["Artwork"]["tmp_name"],
      $path_image . $imageName);
      }
    }else{
    //echo "invalid file";
    }

如何使用自定义名称更改$imageName = $_FILES['Artwork']['name'];，但在名称中保留文件扩展名，例如：myCustomName.jpg

谢谢!

Answer 1

您需要在代码中修改的唯一行是：

$imageName = 'CustomName.' . pathinfo($_FILES['Artwork']['name'],PATHINFO_EXTENSION);

“CustomName”在哪里？是您想要为图像指定的新名称。如果pathinfo函数处理路径和文件名的操作，则为PHP。

你的整个代码应该是：

if ((($_FILES["Artwork"]["type"] == "image/gif")
|| ($_FILES["Artwork"]["type"] == "image/jpeg")
|| ($_FILES["Artwork"]["type"] == "image/jpg")
|| ($_FILES["Artwork"]["type"] == "image/pjpeg"))
&& ($_FILES["Artwork"]["size"] < 20000000))
  {
  if ($_FILES["Artwork"]["error"] > 0)
    {
    //echo "Return Code: " . $_FILES["Artwork"]["error"] . "<br />";
    }else{
      $imageName = 'CustomName.' . pathinfo($_FILES['Artwork']['name'],PATHINFO_EXTENSION);
      move_uploaded_file($_FILES["Artwork"]["tmp_name"],
      $path_image . $imageName);
      }
    }else{
    //echo "invalid file";
    }

Answer 2

$ext = last(explode('.', $_FILES['Artwork']['name']));
$custom_name = 'something';
$imageName = $custom_name.'.'.$ext;

Answer 3

我觉得你把事情搞得太复杂了。只需用点分割文件名并使用最后一个元素即可：

$parts = explode('.', $_FILES['Artwork']['name']);
$newname = "myCustomName" . (size($parts) > 1 ? '.' . last($parts) : '')