本文介绍了我怎么知道一个表是否是一个数组?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在开发一个简单的优化的 JSON 函数. Lua使用表来表示数组,但是在JSON中我需要识别它们之间.使用以下代码:
I'm developing a simple optimized JSON function. Lua uses tables to represent arrays but in JSON I need to recognize between them. The code below is used:
t={
a="hi",
b=100
}
function table2json(t,formatted)
if type(t)~="table" then return nil,"Parameter is not a table. It is: "..type(t) end
local ret=""--return value
local lvl=0 --indentation level
local INDENT=" " --OPTION: the characters put in front of every line for indentation
function addToRet(str) if formatted then ret=ret..string.rep(INDENT,lvl)..str.."\n" else ret=ret..str end end
addToRet("{")
lvl=1
for k,v in pairs(t) do
local typeof=type(v)
if typeof=="string" then
addToRet(k..":\""..v.."\"")
elseif typeof=="number" then
addToRet(k..":"..v)
end
end
lvl=0
addToRet("}")
return ret
end
print(table2json(t,true))
您可以在JSON参考中看到object
在Lua中称为table
,它与array
不同.
As you can see in JSON reference an object
is what is called a table
in Lua and it's different from an array
.
问题是如何检测表是否被用作数组?
The question is how I can detect if a table is being used as an array?
- 当然,一种解决方案是遍历所有对,看看它们是否只有数字连续键,但这不够快.
- 另一种解决方案是在表中放置一个标志,说它是一个数组而不是对象.
有没有更简单/更智能的解决方案?
Any simpler/smarter solution?
推荐答案
谢谢.我开发了以下代码,它可以正常工作:
Thanks. I developed the following code and it works:
---Checks if a table is used as an array. That is: the keys start with one and are sequential numbers
-- @param t table
-- @return nil,error string if t is not a table
-- @return true/false if t is an array/isn't an array
-- NOTE: it returns true for an empty table
function isArray(t)
if type(t)~="table" then return nil,"Argument is not a table! It is: "..type(t) end
--check if all the table keys are numerical and count their number
local count=0
for k,v in pairs(t) do
if type(k)~="number" then return false else count=count+1 end
end
--all keys are numerical. now let's see if they are sequential and start with 1
for i=1,count do
--Hint: the VALUE might be "nil", in that case "not t[i]" isn't enough, that's why we check the type
if not t[i] and type(t[i])~="nil" then return false end
end
return true
end
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