我需要发布以下请求:
POST http://target-host.com/some/endpoint HTTP/1.1
Content-Type: multipart/form-data; boundary="2e3956ac-de47-4cad-90df-05199a7c1f53"
Accept-Encoding: gzip, deflate
Connection: Keep-Alive
Content-Length: 6971
Host: target-host.com
--2e3956ac-de47-4cad-90df-05199a7c1f53
Content-Disposition: form-data; name="some-label"
value
--2e3956ac-de47-4cad-90df-05199a7c1f53
Content-Disposition: form-data; name="file"; filename="my-filename.txt"
<file contents>
--2e3956ac-de47-4cad-90df-05199a7c1f53--
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我可以使用 Pythonrequests库轻松完成此操作,如下所示:
import requests
with open("some_file", "rb") as f:
byte_string = f.read()
requests.post(
"http://target-host.com/some/endpoint",
data={"some-label": "value"},
files={"file": ("my-filename.txt", byte_string)})
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有没有办法对Flurl.Http图书馆做同样的事情?
我的文档记录方式的问题是它将Content-Type为每个键值对插入filename*=utf-8''标头,并为文件数据插入标头。但是,我尝试向其发布请求的服务器不支持此功能。还要注意标题中name和filename值周围的双引号。
编辑:下面是我用来发出帖子请求的代码Flurl.Http:
using System.IO;
using Flurl;
using Flurl.Http;
namespace ConsoleApplication
{
public class Program
{ …Run Code Online (Sandbox Code Playgroud) 我想做类似以下的事情,我想知道它是否与lua完全一致?
fun = function()
some_table = {1,2,3}
-- some dark magic here
return 1,2,3
end
a, b, c = fun()
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如何转换{1,2,3}到1,2,3从功能刚刚恢复?
谢谢
这个问题不是关于读取lua文件,而是读取lua中只读的非可执行文件.
我有一个以下文件
10 awesome % ls -l /sys/class/backlight/intel_backlight/actual_brightness
-r--r--r-- 1 root root 4096 Dec 28 20:17 /sys/class/backlight/intel_backlight/actual_brightness
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但是当我尝试使用以下代码在lua中读取它时,我得到一个错误,即权限被拒绝.有人可以解释一下为什么吗?如何在不更改权限的情况下读取文件?
file = io.popen('/sys/class/backlight/intel_backlight/actual_brightness', 'r')
output = file:read('*all')
file:close()
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错误消息(与lua shell中输入的代码一起:
12 awesome % lua
Lua 5.2.1 Copyright (C) 1994-2012 Lua.org, PUC-Rio
> file = assert(io.popen('/sys/class/backlight/intel_backlight/actual_brightness', 'r'))
> sh: /sys/class/backlight/intel_backlight/actual_brightness: Permission denied
^C
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