连接所有值并将它们附加到同一个变量答案

【问题标题】：Concatenate all values and append them to the same variable连接所有值并将它们附加到同一个变量
【发布时间】：2013-01-01 06:23:41
【问题描述】：

我正在尝试连接所有值并将它们附加到同一个变量，然后写入字符串的内容。我不断收到错误消息：“尝试连接全局 'arglist'（一个 nil 值）”

  eo_device = {}

  eo_device[1] = {node = "ME", t_object = "AV1", h_object = "AV2" }
  eo_device[2] = {node = "ME", t_object = "AV3", h_object = "AV4" }
  eo_device[3] = {node = "ME", t_object = "AV5", h_object = "AV6" }
  eo_device[4] = {node = "ME", t_object = "AV7", h_object = "AV8" }
  eo_device[5] = {node = "ME", t_object = "AV9", h_object = "AV10" }
  eo_device[6] = {node = "ME", t_object = "AV11", h_object = "AV12" }
  eo_device[7] = {node = "ME", t_object = "AV13", h_object = "AV14" }
  eo_device[8] = {node = "ME", t_object = "AV15", h_object = "AV16" }
  eo_device[9] = {node = "ME", t_object = "AV17", h_object = "AV18" }
  eo_device[10] = {node = "ME", t_object = "AV19", h_object = "AV20" }
  eo_device[11] = {node = "ME", t_object = "AV21", h_object = "AV22" }
  eo_device[12] = {node = "ME", t_object = "AV23", h_object = "AV24" }
  eo_device[13] = {node = "ME", t_object = "AV25", h_object = "AV26" }
  eo_device[14] = {node = "ME", t_object = "AV27", h_object = "AV28" }
  eo_device[15] = {node = "ME", t_object = "AV29", h_object = "AV30" }


-- Generate string with all arguments to be written to CSV
for deviceindex = 1,#eo_device do

 device_name = scl.nodes[eo_device[deviceindex].node][eo_device[deviceindex].t_object .. "_Object_Name"].value -- point to Object Names
 description = scl.nodes[eo_device[deviceindex].node][eo_device[deviceindex].t_object .. "_Description"].value -- point to Description of all objects     
 temperature = scl.nodes[eo_device[deviceindex].node][eo_device[deviceindex].t_object .. "_Present_Value"].value -- value of temperature object
 humidity = scl.nodes[eo_device[deviceindex].node][eo_device[deviceindex].h_object .. "_Present_Value"].value -- value of humidity object


 buflist = device_name .. "," .. description .. "," .. temperature .. "," .. humidity 

 arglist = arglist .. "," .. buflist  

 end -- end for

--then once outside of the loop:
arglist = arglist .. "," .. buflist 

buffer = os.date() .. arglist .. "\r\n"

file, info = io.open("trend.csv", "a+") -- write to CSV
if file then
file:write(buffer)
file:close()
end -- end if

【问题讨论】：

标签： csv lua append concatenation

【解决方案1】：

我不断收到错误消息：“尝试连接全局 'arglist'（一个 nil 值）”

因为您试图在代码的第 31 行连接 arglist（一个 nil 值）：

arglist = arglist .. "," .. buflist

arglist 从未初始化，因此当您尝试在= 的右侧进行连接时，您会收到该错误。

关于这个循环的旁注：

for deviceindex = 1,#eo_device do

    description = scl.nodes[eo_device[deviceindex].node][eo_device[deviceindex].t_object .. "_Description"].value

Lua 有一个“通用”for 循环，它允许您遍历 eo_device 的所有元素，为您提供索引和值，因此您不必在您的环形。这将清理您的代码：

for deviceindex, device in ipairs(eo_device) do

    description = scl.nodes[device.node][device.t_object .. "_Description"].value

关于此代码的旁注：

buflist = device_name .. "," .. description .. "," .. temperature .. "," .. humidity

像这样构建字符串效率极低。在循环中执行此操作将导致数千个字符串连接并产生大量垃圾。更好的方法是将所有字符串部分添加到一个表中，然后通过table.concat 将它们连接成一个以逗号分隔的字符串。

例如，要创建一个以逗号分隔的字符串，其中包含 eo_device 表中的所有字符串：

local buffer = {}

for i,device in ipairs(eo_device) do
    table.insert(buffer, device.node)
    table.insert(buffer, device.t_object)
    table.insert(buffer, device.h_object)
end

devices = table.concat(buffer, ',')

【讨论】：

另一个问题，如果我在循环之后 print(bufliststring)，我最终只输出设备 15，这是为什么呢？
我更新了帖子的第 3 部分以包含一个完整的工作示例（缺少您的 eo_device 表）。
hmmm 在打印之前：设备 #15 - Temp,Room #15,74.624,39.2
现在，它打印 ME,AV1,AV2,ME,AV3,AV4,ME,AV5,AV6,ME,AV7,AV8,ME,AV9,AV10,ME,AV11,AV12,ME, AV13,AV14,ME,AV15,AV16,ME,AV17,AV18,ME,AV19,AV20,ME,AV21,AV22,ME,AV23,AV24,ME,AV25,AV26,ME,AV27,AV28,ME,AV29, AV30
所以，我将 table.insert(buffer, device.node) 更改为 table.insert(buffer, device.device_name) 我认为这会有所帮助，但只是打印空白