用Wget发布请求？

Question

我想使用wget将图片上传到远程服务器,使用身份验证令牌'AUTH_1624582364932749DFHDD'到'test'文件夹.

此命令不起作用(授权失败),我想确保它与语法无关:

wget --post-file=nature.jpg http://ipadress:8080/v1/AUTH_test/test/ --post-data="AUTH_1624582364932749DFHDD"

有什么建议？

Answer 1

--post-file 说:

只应指定--post-data和--post-file中的一个.

整段是:

 --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.

特别是它说:

--post-file不用于将文件作为表单附件传输:这些文件必须显示为"key = value"数据

编辑: 您必须了解"key = value"数据的原理.在POST请求中,如在GET请求中,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.它与变量类似.

因此,您不能只是向服务器发送魔术令牌,还需要指定密钥的名称.如果密钥是"令牌",那么它应该是key=value&otherkey=example.

此外,如果可以,您应该考虑使用curl,因为使用它更容易发送文件.互联网上有很多例子.