Question

我使用lua 5.1和luaSocket 2.0.2-4从Web服务器检索页面。我首先检查服务器是否正在响应，然后将Web服务器响应分配给lua变量。

local mysocket = require("socket.http")
if mysocket.request(URL) == nil then
    print('The server is unreachable on:\n'..URL)
    return
end
local response, httpCode, header = mysocket.request(URL)

一切都按预期工作，但请求执行两次。我想知道我是否可以做类似的事情（显然不起作用）：

local mysocket = require("socket.http")
if (local response, httpCode, header = mysocket.request(URL)) == nil then
    print('The server is unreachable on:\n'..URL)
    return
end

Answer 1

是的，就像这样：

local mysocket = require("socket.http")
local response, httpCode, header = mysocket.request(URL)

if response == nil then
    print('The server is unreachable on:\n'..URL)
    return
end

-- here you do your stuff that's supposed to happen when request worked

请求只会发送一次，如果失败，函数将退出。

Answer 2

更好的是，当请求失败时，第二次返回是原因：

如果失败，该函数返回nil，后跟一条错误消息。

（来自the documentation for http.request）

所以你可以直接从插座口打印问题：

local http = require("socket.http")
local response, httpCode, header = http.request(URL)

if response == nil then
    -- the httpCode variable contains the error message instead
    print(httpCode)
    return
end

-- here you do your stuff that's supposed to happen when request worked

Lua http socket评估

2 个答案: