Question

我正在使用此功能上传我的文件：

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一一行是：

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

'CustomName'的位置。是图像所需的新名称。 pathinfo如果PHP函数用路径和文件名来处理操作。

您的整个代码将是：

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

我认为你太复杂了。它就像用点分割文件名并使用最后一个元素一样简单：

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

PHP上传文件并提供自定义名称

3 个答案: