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decode 会给空的json对象添加元表 ‘json.object’, 导致table.insert(t), encode 产生错误的结果 about lua-rapidjson HOT 1 CLOSED

sniper00 avatar sniper00 commented on July 2, 2024
decode 会给空的json对象添加元表 ‘json.object’, 导致table.insert(t), encode 产生错误的结果

from lua-rapidjson.

Comments (1)

xpol avatar xpol commented on July 2, 2024

This should work:

local json = require("json")
local t = {a={},b={}}
local jsonstr = json.encode(t, {empty_table_as_array = true})
local tb = json.decode(jsonstr)
table.insert(tb.a,"hello world")

from lua-rapidjson.

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