我在显示包含嵌套表（n-deep）的表的内容时遇到问题。我想通过print语句或快速而肮脏的东西将它转储到std out或控制台，但我无法弄清楚如何。我正在寻找使用gdb打印NSDictionary时得到的粗略等价物。
答案 0 :(得分：82)
我知道这个问题已被标记为已回答，但让我在这里插入自己的库。它叫做inspect.lua，你可以在这里找到它：
https://github.com/kikito/inspect.lua
它只是一个文件，您可以从任何其他文件中获取。它返回一个函数，将任何Lua值转换为人类可读的字符串：
local inspect = require('inspect')
print(inspect({1,2,3})) -- {1, 2, 3}
print(inspect({a=1,b=2})
-- {
-- a = 1
-- b = 2
-- }
它正确缩进子表，并正确处理“递归表”（包含对自身的引用的表），因此它不会进入无限循环。它以合理的方式对价值进行排序。它还可以打印元信息。
问候！
答案 1 :(得分：47)
我发现这个有用。因为如果递归它也可以打印嵌套表。
function dump(o)
if type(o) == 'table' then
local s = '{ '
for k,v in pairs(o) do
if type(k) ~= 'number' then k = '"'..k..'"' end
s = s .. '['..k..'] = ' .. dump(v) .. ','
end
return s .. '} '
else
return tostring(o)
end
end
e.g。
local people = {
{
name = "Fred",
address = "16 Long Street",
phone = "123456"
},
{
name = "Wilma",
address = "16 Long Street",
phone = "123456"
},
{
name = "Barney",
address = "17 Long Street",
phone = "123457"
}
}
print("People:", dump(people))
产生以下输出：
人物：{[1] = {[＆＃34;地址＆＃34;] = 16长街，[＆＃34;电话＆＃34;] = 123456，[＆＃34; name＆＃34;] = Fred，}，[2] = {[＆＃34;地址＆＃34;] = 16长 Street，[＆＃34; phone＆＃34;] = 123456，[＆＃34; name＆＃34;] = Wilma，}，[3] = {[＆＃34;地址＆＃34;] = 17 Long Street，[＆＃34; phone＆＃34;] = 123457，[＆＃34; name＆＃34;] = Barney，}，}
答案 2 :(得分：46)
随意浏览Lua Wiki on table serialization。它列出了有关如何将表转储到控制台的几种方法。
你必须选择哪一个最适合你。有很多方法可以做到，但我通常最终使用Penlight中的那个：
> t = { a = { b = { c = "Hello world!", 1 }, 2, d = { 3 } } }
> require 'pl.pretty'.dump(t)
{
a = {
d = {
3
},
b = {
c = "Hello world!",
1
},
2
}
}
答案 3 :(得分：17)
发现了这个：
-- Print contents of `tbl`, with indentation.
-- `indent` sets the initial level of indentation.
function tprint (tbl, indent)
if not indent then indent = 0 end
for k, v in pairs(tbl) do
formatting = string.rep(" ", indent) .. k .. ": "
if type(v) == "table" then
print(formatting)
tprint(v, indent+1)
elseif type(v) == 'boolean' then
print(formatting .. tostring(v))
else
print(formatting .. v)
end
end
end
从这里开始 https://gist.github.com/ripter/4270799
对我来说非常好......
答案 4 :(得分：8)
我见过的大多数纯lua打印表函数都存在深度递归问题 当走得太深时往往会导致堆栈溢出。这个印刷品 我写过的表函数没有这个问题。由于它处理串联的方式，它还应该能够处理非常大的表。在我个人使用这个功能时，它在大约一秒内输出了63k行文件。
输出还保持lua语法，并且可以轻松修改脚本 通过将输出写入文件（如果修改为允许）来实现简单的持久存储 只有要格式化的数字，布尔值，字符串和表数据类型。
function print_table(node)
-- to make output beautiful
local function tab(amt)
local str = ""
for i=1,amt do
str = str .. "\t"
end
return str
end
local cache, stack, output = {},{},{}
local depth = 1
local output_str = "{\n"
while true do
local size = 0
for k,v in pairs(node) do
size = size + 1
end
local cur_index = 1
for k,v in pairs(node) do
if (cache[node] == nil) or (cur_index >= cache[node]) then
if (string.find(output_str,"}",output_str:len())) then
output_str = output_str .. ",\n"
elseif not (string.find(output_str,"\n",output_str:len())) then
output_str = output_str .. "\n"
end
-- This is necessary for working with HUGE tables otherwise we run out of memory using concat on huge strings
table.insert(output,output_str)
output_str = ""
local key
if (type(k) == "number" or type(k) == "boolean") then
key = "["..tostring(k).."]"
else
key = "['"..tostring(k).."']"
end
if (type(v) == "number" or type(v) == "boolean") then
output_str = output_str .. tab(depth) .. key .. " = "..tostring(v)
elseif (type(v) == "table") then
output_str = output_str .. tab(depth) .. key .. " = {\n"
table.insert(stack,node)
table.insert(stack,v)
cache[node] = cur_index+1
break
else
output_str = output_str .. tab(depth) .. key .. " = '"..tostring(v).."'"
end
if (cur_index == size) then
output_str = output_str .. "\n" .. tab(depth-1) .. "}"
else
output_str = output_str .. ","
end
else
-- close the table
if (cur_index == size) then
output_str = output_str .. "\n" .. tab(depth-1) .. "}"
end
end
cur_index = cur_index + 1
end
if (size == 0) then
output_str = output_str .. "\n" .. tab(depth-1) .. "}"
end
if (#stack > 0) then
node = stack[#stack]
stack[#stack] = nil
depth = cache[node] == nil and depth + 1 or depth - 1
else
break
end
end
-- This is necessary for working with HUGE tables otherwise we run out of memory using concat on huge strings
table.insert(output,output_str)
output_str = table.concat(output)
print(output_str)
end
以下是一个例子：
local t = {
["abe"] = {1,2,3,4,5},
"string1",
50,
["depth1"] = { ["depth2"] = { ["depth3"] = { ["depth4"] = { ["depth5"] = { ["depth6"] = { ["depth7"]= { ["depth8"] = { ["depth9"] = { ["depth10"] = {1000}, 900}, 800},700},600},500}, 400 }, 300}, 200}, 100},
["ted"] = {true,false,"some text"},
"string2",
[function() return end] = function() return end,
75
}
print_table(t)
输出：
{
[1] = 'string1',
[2] = 50,
[3] = 'string2',
[4] = 75,
['abe'] = {
[1] = 1,
[2] = 2,
[3] = 3,
[4] = 4,
[5] = 5
},
['function: 06472B70'] = 'function: 06472A98',
['depth1'] = {
[1] = 100,
['depth2'] = {
[1] = 200,
['depth3'] = {
[1] = 300,
['depth4'] = {
[1] = 400,
['depth5'] = {
[1] = 500,
['depth6'] = {
[1] = 600,
['depth7'] = {
[1] = 700,
['depth8'] = {
[1] = 800,
['depth9'] = {
[1] = 900,
['depth10'] = {
[1] = 1000
}
}
}
}
}
}
}
}
}
},
['ted'] = {
[1] = true,
[2] = false,
[3] = 'some text'
}
}
答案 5 :(得分：4)
如前所述，你必须写它。 这是我的简陋版本:(超级基本版）
function tprint (t, s)
for k, v in pairs(t) do
local kfmt = '["' .. tostring(k) ..'"]'
if type(k) ~= 'string' then
kfmt = '[' .. k .. ']'
end
local vfmt = '"'.. tostring(v) ..'"'
if type(v) == 'table' then
tprint(v, (s or '')..kfmt)
else
if type(v) ~= 'string' then
vfmt = tostring(v)
end
print(type(t)..(s or '')..kfmt..' = '..vfmt)
end
end
end
示例：
local mytbl = { ['1']="a", 2, 3, b="c", t={d=1} }
tprint(mytbl)
输出（Lua 5.0）：
table[1] = 2
table[2] = 3
table["1"] = "a"
table["t"]["d"] = 1
table["b"] = "c"
答案 6 :(得分：2)
metalua 的table.tostring metehod实际上非常完整。它处理嵌套表，缩进级别是可变的，...
见https://github.com/fab13n/metalua/blob/master/src/lib/metalua/table2.lua
答案 7 :(得分：2)
这是我的版本，支持排除表格和用户数据
-- Lua Table View by Elertan
table.print = function(t, exclusions)
local nests = 0
if not exclusions then exclusions = {} end
local recurse = function(t, recurse, exclusions)
indent = function()
for i = 1, nests do
io.write(" ")
end
end
local excluded = function(key)
for k,v in pairs(exclusions) do
if v == key then
return true
end
end
return false
end
local isFirst = true
for k,v in pairs(t) do
if isFirst then
indent()
print("|")
isFirst = false
end
if type(v) == "table" and not excluded(k) then
indent()
print("|-> "..k..": "..type(v))
nests = nests + 1
recurse(v, recurse, exclusions)
elseif excluded(k) then
indent()
print("|-> "..k..": "..type(v))
elseif type(v) == "userdata" or type(v) == "function" then
indent()
print("|-> "..k..": "..type(v))
elseif type(v) == "string" then
indent()
print("|-> "..k..": ".."\""..v.."\"")
else
indent()
print("|-> "..k..": "..v)
end
end
nests = nests - 1
end
nests = 0
print("### START TABLE ###")
for k,v in pairs(t) do
print("root")
if type(v) == "table" then
print("|-> "..k..": "..type(v))
nests = nests + 1
recurse(v, recurse, exclusions)
elseif type(v) == "userdata" or type(v) == "function" then
print("|-> "..k..": "..type(v))
elseif type(v) == "string" then
print("|-> "..k..": ".."\""..v.."\"")
else
print("|-> "..k..": "..v)
end
end
print("### END TABLE ###")
end
这是一个例子
t = {
location = {
x = 10,
y = 20
},
size = {
width = 100000000,
height = 1000,
},
name = "Sidney",
test = {
hi = "lol",
},
anotherone = {
1,
2,
3
}
}
table.print(t, { "test" })
打印：
### START TABLE ###
root
|-> size: table
|
|-> height: 1000
|-> width: 100000000
root
|-> location: table
|
|-> y: 20
|-> x: 10
root
|-> anotherone: table
|
|-> 1: 1
|-> 2: 2
|-> 3: 3
root
|-> test: table
|
|-> hi: "lol"
root
|-> name: "Sidney"
### END TABLE ###
请注意，根目录不会删除排除
答案 8 :(得分：1)
你必须自己编码我害怕。我写了这篇文章，它可能会对你有所帮助
function printtable(table, indent)
indent = indent or 0;
local keys = {};
for k in pairs(table) do
keys[#keys+1] = k;
table.sort(keys, function(a, b)
local ta, tb = type(a), type(b);
if (ta ~= tb) then
return ta < tb;
else
return a < b;
end
end);
end
print(string.rep(' ', indent)..'{');
indent = indent + 1;
for k, v in pairs(table) do
local key = k;
if (type(key) == 'string') then
if not (string.match(key, '^[A-Za-z_][0-9A-Za-z_]*$')) then
key = "['"..key.."']";
end
elseif (type(key) == 'number') then
key = "["..key.."]";
end
if (type(v) == 'table') then
if (next(v)) then
printf("%s%s =", string.rep(' ', indent), tostring(key));
printtable(v, indent);
else
printf("%s%s = {},", string.rep(' ', indent), tostring(key));
end
elseif (type(v) == 'string') then
printf("%s%s = %s,", string.rep(' ', indent), tostring(key), "'"..v.."'");
else
printf("%s%s = %s,", string.rep(' ', indent), tostring(key), tostring(v));
end
end
indent = indent - 1;
print(string.rep(' ', indent)..'}');
end
答案 9 :(得分：1)
--~ print a table
function printTable(list, i)
local listString = ''
--~ begin of the list so write the {
if not i then
listString = listString .. '{'
end
i = i or 1
local element = list[i]
--~ it may be the end of the list
if not element then
return listString .. '}'
end
--~ if the element is a list too call it recursively
if(type(element) == 'table') then
listString = listString .. printTable(element)
else
listString = listString .. element
end
return listString .. ', ' .. printTable(list, i + 1)
end
local table = {1, 2, 3, 4, 5, {'a', 'b'}, {'G', 'F'}}
print(printTable(table))
嗨，伙计，我写了一个siple代码，在纯Lua中执行此操作，它有一个错误（在列表的最后一个元素之后写一个昏迷）但是我如何快速编写它作为原型我会让它适应你它符合您的需求。
答案 10 :(得分：1)
格式为 JSON （您可以＆＃34;美化＆＃34;稍后在IDE中）：
let data = {
"message": "The given data was invalid.",
"errors": {
"roomtype_id": [
"Please select a room type"
],
"price": [
"Price cannot be empty"
],
"amenities": [
"Atleast select one amenities"
],
"floor.0": [
"Please select a floor"
],
"floor.1": [
"Please select a floor"
],
"floor.2": [
"Please select a floor"
],
"floor.3": [
"Please select a floor"
],
"floor.4": [
"Please select a floor"
],
"room_number.0": [
"Room number cannot be empty"
],
"room_number.1": [
"Room number cannot be empty"
],
"room_number.2": [
"Room number cannot be empty"
],
"room_number.3": [
"Room number cannot be empty"
],
"room_number.4": [
"Room number cannot be empty"
],
"room_number.5": [
"Room number cannot be empty"
],
"room_number.6": [
"Room number cannot be empty"
],
"room_number.7": [
"Room number cannot be empty"
],
"room_number.8": [
"Room number cannot be empty"
],
"room_number.9": [
"Room number cannot be empty"
],
"room_number.10": [
"Room number cannot be empty"
],
"room_number.11": [
"Room number cannot be empty"
],
"room_number.12": [
"Room number cannot be empty"
],
"room_number.13": [
"Room number cannot be empty"
],
"room_number.14": [
"Room number cannot be empty"
],
"room_number.15": [
"Room number cannot be empty"
],
"room_number.16": [
"Room number cannot be empty"
],
"room_number.17": [
"Room number cannot be empty"
],
"room_number.18": [
"Room number cannot be empty"
],
"room_number.19": [
"Room number cannot be empty"
],
"room_number.20": [
"Room number cannot be empty"
],
"room_number.21": [
"Room number cannot be empty"
],
"room_number.22": [
"Room number cannot be empty"
],
"room_number.23": [
"Room number cannot be empty"
],
"room_number.24": [
"Room number cannot be empty"
],
"room_number.25": [
"Room number cannot be empty"
],
"room_number.26": [
"Room number cannot be empty"
],
"room_number.27": [
"Room number cannot be empty"
],
"room_number.28": [
"Room number cannot be empty"
]
}
}
let errorKeys = Object.keys(data.errors).filter(x => x.startsWith('floor'));
let count = errorKeys.length;
let floorErrors = errorKeys.map(key => ({
key: key,
errors: data.errors[key]
}));
console.log(errorKeys, count, floorErrors)
console.log()用法：
local function format_any_value(obj, buffer)
local _type = type(obj)
if _type == "table" then
buffer[#buffer + 1] = '{"'
for key, value in next, obj, nil do
buffer[#buffer + 1] = tostring(key) .. '":'
format_any_value(value, buffer)
buffer[#buffer + 1] = ',"'
end
buffer[#buffer] = '}' -- note the overwrite
elseif _type == "string" then
buffer[#buffer + 1] = '"' .. obj .. '"'
elseif _type == "boolean" or _type == "number" then
buffer[#buffer + 1] = tostring(obj)
else
buffer[#buffer + 1] = '"???' .. _type .. '???"'
end
end
BTW，我还写了几个其他解决方案：非常快一个，还有一个特殊字符转义：https://github.com/vn971/fast_json_encode
答案 11 :(得分：1)
最简单的方法，具有循环引用处理和所有功能：
function dump(t, indent, done)
done = done or {}
indent = indent or 0
done[t] = true
for key, value in pairs(t) do
print(string.rep("\t", indent))
if (type(value) == "table" and not done[value]) then
done[value] = true
print(key, ":\n")
dump(value, indent + 2, done)
done[value] = nil
else
print(key, "\t=\t", value, "\n")
end
end
end
答案 12 :(得分：0)
我使用自己的函数来打印表的内容，但不确定如何将其转换为您的环境：
---A helper function to print a table's contents.
---@param tbl table @The table to print.
---@param depth number @The depth of sub-tables to traverse through and print.
---@param n number @Do NOT manually set this. This controls formatting through recursion.
function Lib:PrintTable(tbl, depth, n)
n = n or 0;
depth = depth or 5;
if (depth == 0) then
print(string.rep(' ', n).."...");
return;
end
if (n == 0) then
print(" ");
end
for key, value in pairs(tbl) do
if (key and self:IsNumber(key) or self:IsString(key)) then
key = string.format("[\"%s\"]", key);
if (self:IsTable(value)) then
if (next(value)) then
print(string.rep(' ', n)..key.." = {");
self:PrintTable(value, depth - 1, n + 4);
print(string.rep(' ', n).."},");
else
print(string.rep(' ', n)..key.." = {},");
end
else
if (self:IsString(value)) then
value = string.format("\"%s\"", value);
else
value = tostring(value);
end
print(string.rep(' ', n)..key.." = "..value..",");
end
end
end
if (n == 0) then
print(" ");
end
end
答案 13 :(得分：0)
添加另一个版本。该尝试也可以遍历用户数据。
function inspect(o,indent)
if indent == nil then indent = 0 end
local indent_str = string.rep(" ", indent)
local output_it = function(str)
print(indent_str..str)
end
local length = 0
local fu = function(k, v)
length = length + 1
if type(v) == "userdata" or type(v) == 'table' then
output_it(indent_str.."["..k.."]")
inspect(v, indent+1)
else
output_it(indent_str.."["..k.."] "..tostring(v))
end
end
local loop_pairs = function()
for k,v in pairs(o) do fu(k,v) end
end
local loop_metatable_pairs = function()
for k,v in pairs(getmetatable(o)) do fu(k,v) end
end
if not pcall(loop_pairs) and not pcall(loop_metatable_pairs) then
output_it(indent_str.."[[??]]")
else
if length == 0 then
output_it(indent_str.."{}")
end
end
end
答案 14 :(得分：0)
lua 转储表的简单例子
我建议使用 serpent.lua
local function parser(value, indent, subcategory)
local indent = indent or 2
local response = '(\n'
local subcategory = type(subcategory) == 'number' and subcategory or indent
for key, value in pairs(value) do
if type(value) == 'table' then
value = parser(value, indent, subcategory + indent)
elseif type(value) == 'string' then
value = '\''.. value .. '\''
elseif type(value) ~= 'number' then
value = tostring(value)
end
if type(tonumber(key)) == 'number' then
key = '[' .. key .. ']'
elseif not key:match('^([A-Za-z_][A-Za-z0-9_]*)$') then
key = '[\'' .. key .. '\']'
end
response = response .. string.rep(' ', subcategory) .. key .. ' = ' .. value .. ',\n'
end
return response .. string.rep(' ', subcategory - indent) .. ')'
end
示例
response = parser{1,2,3, {ok = 10, {}}}
print(response)
结果
(
[1] = 1,
[2] = 2,
[3] = 3,
[4] = (
[1] = (),
ok = 10
)
)
答案 15 :(得分：-1)
我谦虚地修改了一些Alundaio代码：
-- by Alundaio
-- KK modified 11/28/2019
function dump_table_to_string(node, tree, indentation)
local cache, stack, output = {},{},{}
local depth = 1
if type(node) ~= "table" then
return "only table type is supported, got " .. type(node)
end
if nil == indentation then indentation = 1 end
local NEW_LINE = "\n"
local TAB_CHAR = " "
if nil == tree then
NEW_LINE = "\n"
elseif not tree then
NEW_LINE = ""
TAB_CHAR = ""
end
local output_str = "{" .. NEW_LINE
while true do
local size = 0
for k,v in pairs(node) do
size = size + 1
end
local cur_index = 1
for k,v in pairs(node) do
if (cache[node] == nil) or (cur_index >= cache[node]) then
if (string.find(output_str,"}",output_str:len())) then
output_str = output_str .. "," .. NEW_LINE
elseif not (string.find(output_str,NEW_LINE,output_str:len())) then
output_str = output_str .. NEW_LINE
end
-- This is necessary for working with HUGE tables otherwise we run out of memory using concat on huge strings
table.insert(output,output_str)
output_str = ""
local key
if (type(k) == "number" or type(k) == "boolean") then
key = "["..tostring(k).."]"
else
key = "['"..tostring(k).."']"
end
if (type(v) == "number" or type(v) == "boolean") then
output_str = output_str .. string.rep(TAB_CHAR,depth*indentation) .. key .. " = "..tostring(v)
elseif (type(v) == "table") then
output_str = output_str .. string.rep(TAB_CHAR,depth*indentation) .. key .. " = {" .. NEW_LINE
table.insert(stack,node)
table.insert(stack,v)
cache[node] = cur_index+1
break
else
output_str = output_str .. string.rep(TAB_CHAR,depth*indentation) .. key .. " = '"..tostring(v).."'"
end
if (cur_index == size) then
output_str = output_str .. NEW_LINE .. string.rep(TAB_CHAR,(depth-1)*indentation) .. "}"
else
output_str = output_str .. ","
end
else
-- close the table
if (cur_index == size) then
output_str = output_str .. NEW_LINE .. string.rep(TAB_CHAR,(depth-1)*indentation) .. "}"
end
end
cur_index = cur_index + 1
end
if (size == 0) then
output_str = output_str .. NEW_LINE .. string.rep(TAB_CHAR,(depth-1)*indentation) .. "}"
end
if (#stack > 0) then
node = stack[#stack]
stack[#stack] = nil
depth = cache[node] == nil and depth + 1 or depth - 1
else
break
end
end
-- This is necessary for working with HUGE tables otherwise we run out of memory using concat on huge strings
table.insert(output,output_str)
output_str = table.concat(output)
return output_str
end
然后：
print(dump_table_to_string("AA", true,3))
print(dump_table_to_string({"AA","BB"}, true,3))
print(dump_table_to_string({"AA","BB"}))
print(dump_table_to_string({"AA","BB"},false))
print(dump_table_to_string({"AA","BB",{22,33}},true,2))
给予：
only table type is supported, got string
{
[1] = 'AA',
[2] = 'BB'
}
{
[1] = 'AA',
[2] = 'BB'
}
{[1] = 'AA',[2] = 'BB'}
{
[1] = 'AA',
[2] = 'BB',
[3] = {
[1] = 22,
[2] = 33
}
}