Lua如何根据当前日期对表进行排序

每周的每一天都被指定了一个数字，从星期日的 1 开始，到星期六的 7 结束。为了找到当前日期的数字表示，调用 os.date("*t")，它会返回一个包含有关当前日期、月份、年份等信息的表。与当前日期数字相对应的字段称为 wday。

for k, v in pairs(os.date("*t")) do
    print(k, v)
end

输出：

year    2020
wday    6
month   8
isdst   true
hour    21
day 28
sec 13
yday    241
min 48

这里，与键 wday 相关联的值为 6，对应星期五。

有了这个数字，您可以通过弹出当前日期之前的元素，然后将它们重新插入到末尾来重新排序天数表。

-- Local references to table functions.
local tblins = table.insert
local tblrmv = table.remove

local days = {
    "Sunday",
    "Monday",
    "Tuesday",
    "Wednesday",
    "Thursday",
    "Friday",
    "Saturday"
}

local function reorder(days, wday)
    for i = wday - 1, 1, -1 do
        -- Pop element days[1] and then append it.
        tblins(days, tblrmv(days, 1))
    end
    return
end

local date_table = os.date("*t")
reorder(days, date_table.wday)

for i, day in ipairs(days) do
    print(i, day)
end

输出：

1   Friday
2   Saturday
3   Sunday
4   Monday
5   Tuesday
6   Wednesday
7   Thursday

如果您想要一个易于重新启动的版本，这里是我现有解决方案的扩展：

-- Local references to table functions.
local tblins = table.insert
local tblrmv = table.remove
local tblsrt = table.sort

-- The field `day' refers to the string representation; `num' refers to the
-- given day's original position in the table, which is used to restore the
-- table to its starting order.
local days = {
    {day = "Sunday", num = 1},
    {day = "Monday", num = 2},
    {day = "Tuesday", num = 3},
    {day = "Wednesday", num = 4},
    {day = "Thursday", num = 5},
    {day = "Friday", num = 6},
    {day = "Sunday", num = 7}
}

local function sort_days(left_day, right_day)
    return left_day.num < right_day.num
end

-- The reorder function remains the same
-- To put the days table back in its original order, call the following:
tblsrt(days, sort_days)

for i, day in ipairs(days) do
    print(i, day.day, day.num)
end

输出：

1   Sunday  1
2   Monday  2
3   Tuesday 3
4   Wednesday   4
5   Thursday    5
6   Friday  6
7   Saturday    7