如何阅读在PHP上传文件的标题？

问题描述

我们可以在PHP中读取一个文件的头文件信息，以确定上传文件的类型吗？

我不想依赖 $ _ FILES ['control_name_from_client'] ['type']。我们知道这个属性通过读取文件的扩展名来确定文件类型。 b $ b

如果用户重命名 test.jpg - > test.xls ，该怎么办？在这种情况下， $ _ FILES ['control_name_from_client'] ['type'] 将显示类型为 application / vnd.ms-excel 而不是 image / jpeg 。这是很自然的，如果需要执行代码来读取XLS文件来获取数据以进行一些处理，那么这可能会造成问题。

有什么建议吗？

解决方案

试试 即可。你必须调用它传递文件路径。例如：

  $ finfo = finfo_open（FILEINFO_MIME_TYPE）; 
 $ mime = finfo_file（$ finfo，$ _FILES ['control_name_from_client'] ['tmp_name']）; 
 finfo_close（$ finfo）; 
  

您需要 Fileinfo 扩展名。正如PHP手册所说：
$ b

Can we read the header information of a file in PHP to determine the type of file uploaded?.

I don't want to rely on $_FILES['control_name_from_client']['type']. As we know that this property determines the file type by reading the extension of the file uploaded.

What if the user renames, say test.jpg -> test.xls. In that case, $_FILES['control_name_from_client']['type'] will show the type as application/vnd.ms-excel instead of image/jpeg. It is but natural this can create problems if a code has to be executed which reads the XLS file to fetch data for some processing.

Any suggestions please?

解决方案

Try finfo_file(). You have to call it passing the filepath. Example:

$finfo = finfo_open(FILEINFO_MIME_TYPE);
$mime = finfo_file($finfo, $_FILES['control_name_from_client']['tmp_name']);
finfo_close($finfo);

You need the Fileinfo extension. As PHP manual says:

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