当密钥正确时，OpenAI API返回一个“未经授权”的错误。

Question

我在Lua中编写了一个不和谐的机器人，我认为以某种方式实现OpenAI的api会很有趣，除了我一直有401错误之外，我已经做好了所有的事情。这是我的代码的一部分

coroutine.wrap(function()
    local s,e = pcall(function()
        local Headers = {
            ["Authorization"] = "Bearer "..key,
            ["Content-Type"] = "application/json",
        }

        local Body = json.encode({
            model = "text-davinci-002",
            prompt = "Human: ".. table.concat(Args, " ") .. "\n\nAI:",
            temperature = 0.9,
            max_tokens = 47, --150
            top_p = 1,
            frequency_penalty = 0.0,
            presence_penalty = 0.6,
            stop = {" Human:", " AI:"}
        })
    
        res,body = coro.request("POST", link, Headers, Body, 5000)

        if res == nil then
            Message:reply("didnt return anything")
            return
        end
        
        if res.code < 200 or res.code >= 300 then
            Message:reply("Failed to send request: " .. res.reason); return --Always ends up here "Failed to send request: Unauthorized"
        end

        Message:reply("Request sent successfully!")
    end)
end)()

“密钥”是我从网站获得的API密钥。我觉得这个错误是简单而愚蠢的，但不管我被困在哪里

Answer 1

这是很好的代码，不过在您验证代码之前，我会对这些类型做一些检查。

这背后的另一个原因是，某些域可能需要代理而不是直接连接。

coroutine.resume(coroutine.create(function()
    local headers = {
        Authorization = "Bearer " .. key,
        ["Content-Type"] = "application/json",
    }
    local body = json.encode({
        model = "text-davinci-002",
        prompt = "Human: " .. table.concat(Args, " ") .. "\n\nAI:",
        temperature = 0.9,
        max_tokens = 47, --150
        top_p = 1,
        frequency_penalty = 0.0,
        presence_penalty = 0.6,
        stop = { " Human:", " AI:" },
    })
    local success, http_result, http_body = pcall(coro.request, "POST", link, headers, body, 5e3)
    if success ~= true then
        return error(http_result, 0)
    elseif type(http_result) == "table" and type(http_result.code) == "number" and http_result.code < 200 or http_result.code >= 300 then
        return Message:reply("Failed to send request: " .. type(http_result.reason) == "string" and http_result.reason or "No reason provided.")
    end
    return Message:reply("Request sent successfully!")
end))