Wget当前仅支持x-www-form-urlencoded数据。--post- file不是用于将文件作为表单附件传输，而是需要格式为的数据key=value&otherkey=example。

--post-data并--post-file以相同的方式工作：唯一的不同是，--post-data您可以在命令行中指定数据，而--post- file允许您指定包含要发送的数据的文件的路径。

这里是文档：

 --post-data=string
       --post-file=file
           Use POST as the method for all HTTP requests and send the specified data
           in the request body.  --post-data sends string as data, whereas
           --post-file sends the contents of file.  Other than that, they work in
           exactly the same way. In particular, they both expect content of the
           form "key1=value1&key2=value2", with percent-encoding for special
           characters; the only difference is that one expects its content as a
           command-line parameter and the other accepts its content from a file. In
           particular, --post-file is not for transmitting files as form
           attachments: those must appear as "key=value" data (with appropriate
           percent-coding) just like everything else. Wget does not currently
           support "multipart/form-data" for transmitting POST data; only
           "application/x-www-form-urlencoded". Only one of --post-data and
           --post-file should be specified.

关于身份验证令牌，应该在标头，URL路径或数据本身中提供它。必须在您使用的服务的文档中的某处指出。在POST请求中，就像在GET请求中一样，您必须使用键和值指定数据。这样，服务器将能够接收具有特定名称的多个信息。变量也是如此。

因此，您不仅可以将魔术令牌发送到服务器，还需要指定密钥的名称。如果密钥是“令牌”，则应该是token=YOUR_TOKEN。

wget --post-data 'user=foo&password=bar' http://example.com/auth.php

另外，如果可以的话，应该考虑使用curl，因为使用它可以更轻松地发送文件。互联网上有很多例子。