我需要一个Lua 4脚本,将seconds = 0
后经过的秒数转换为D:HH:MM:SS格式的字符串。我查看的方法尝试将数字转换为日历日期和时间,但我只需要自0
以来经过的时间。如果日值增加到数百或数千,那就没问题。我该如何编写这样的脚本?
答案 0 :(得分:4)
试试这个:
function disp_time(time)
local days = floor(time/86400)
local remaining = time % 86400
local hours = floor(remaining/3600)
remaining = remaining % 3600
local minutes = floor(remaining/60)
remaining = remaining % 60
local seconds = remaining
if (hours < 10) then
hours = "0" .. tostring(hours)
end
if (minutes < 10) then
minutes = "0" .. tostring(minutes)
end
if (seconds < 10) then
seconds = "0" .. tostring(seconds)
end
answer = tostring(days)..':'..hours..':'..minutes..':'..seconds
return answer
end
cur_time = os.time()
print(disp_time(cur_time))
答案 1 :(得分:2)
这与其他答案类似,但更短。返回行使用格式字符串以D:HH:MM:SS格式显示结果。
function disp_time(time)
local days = floor(time/86400)
local hours = floor(mod(time, 86400)/3600)
local minutes = floor(mod(time,3600)/60)
local seconds = floor(mod(time,60))
return format("%d:%02d:%02d:%02d",days,hours,minutes,seconds)
end
答案 2 :(得分:0)
我找到了一个能够适应的Java示例。
function seconds_to_days_hours_minutes_seconds(total_seconds)
local time_days = floor(total_seconds / 86400)
local time_hours = floor(mod(total_seconds, 86400) / 3600)
local time_minutes = floor(mod(total_seconds, 3600) / 60)
local time_seconds = floor(mod(total_seconds, 60))
if (time_hours < 10) then
time_hours = "0" .. time_hours
end
if (time_minutes < 10) then
time_minutes = "0" .. time_minutes
end
if (time_seconds < 10) then
time_seconds = "0" .. time_seconds
end
return time_days .. ":" .. time_hours .. ":" .. time_minutes .. ":" .. time_seconds
end