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Formating seconds to string format.

 
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Corroder
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PostPosted: Mon Jul 22, 2019 1:02 am    Post subject: Formating seconds to string format. Reply with quote

I try to output the time span in seconds to following:

Code:
function SecondsFormat(X)
    if X <= 0 then return "" end
    local date = os.date("%c", X)
    local inPattern = "(%d+)/(%d+)/(%d+) (%d+):(%d+):(%d+)"
    local outPattern = "%dy:%dm:%dd:%dh:%dm:%ds"
    local month, day, year, hour, minute, second = date:match(inPattern)
    month = month - 1
    day = day - 1
    year = year - 70
    return string.format(outPattern, year, month, day, hour, minute, second)
end


The function above will output, string format example: 1y:2m:21d:17h:25m:11s
But, I want output as: 01:02:21:17:25:11
How to write the output pattern?

And let say I have variables:

Code:
local year, month, day, hour, minute, second = 2019, 12, 31, 23, 59, 01


Then I need to concatenate those variables so I can use os.time(os.date(*t)) from a specific date+time as store on those variables. How?


Thanks.

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Dark Byte
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PostPosted: Mon Jul 22, 2019 2:13 am    Post subject: Reply with quote

use %.2d for the 2 digit format, no idea about the rest
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AylinCE
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PostPosted: Mon Jul 22, 2019 4:55 am    Post subject: Reply with quote

it seems to give you the result you want.

Code:
function SecondsFormat()
    local year = tostring(os.date('%Y')) -- or %y = 19:07:22-13:49:07
    local month = tostring(os.date('%m'))
    local day = tostring(os.date('%d'))
    local hour = tostring(os.date('%p' < '%I' or '%H'))
    local minute = tostring(os.date('%M'))
    local second = tostring(os.date('%S'))
    print(tostring(os.date(year..':'..month..':'..day..'-'..hour..':'..minute..':'..second)))
end
SecondsFormat()
--result = 2019:07:22-13:50:30



Code:
function SecondsFormat1()
    local year = tostring(os.date('%Y')) -- or %y = 19
    local month = tostring(os.date('%B'))
    local day = tostring(os.date('%A'))
    local hour = tostring(os.date('%p' < '%I' or '%H'))
    local minute = tostring(os.date('%M'))
    local second = tostring(os.date('%S'))
    print(tostring(os.date(year..' / '..month..' / '..day..' /-/ '..hour..':'..minute..':'..second)))
end
SecondsFormat1()

--result: 2019 / July / Monday /-/ 14:03:21

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Corroder
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PostPosted: Mon Jul 22, 2019 5:40 am    Post subject: Reply with quote

Dark Byte wrote:
use %.2d for the 2 digit format, no idea about the rest


Yes, thank DB. "%2d" give 2 digit format value, but not leading zero value if the value as 1 digit. So, I use "%02.f" to output all value with 2 digits.


@Aylin : That not I want. What I want is to return seconds or seconds left to date and time format. I solved it :

Code:
local yr, mn, dy, hr, mt, sc = 2019, 12, 31, 23, 59, 01
local date_tbl = { year = yr, month = mn, day = dy, hour = hr, minute = mt, second = sc }
local targetTime = os.time( date_tbl )

print(os.date("%a, %d %b %Y  %H:%M:%S", targetTime ))
-- result :  Tue, 31 Dec 2019  23:00:00

print(os.date( "!%A, %d-%B-%Y %H:%M:%S" , targetTime + 1 * 60 * 60 ))
-- result :  Tuesday, 31-December-2019 17:00:00

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Dark Byte
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PostPosted: Mon Jul 22, 2019 9:14 am    Post subject: Reply with quote

Corroder wrote:
Dark Byte wrote:
use %.2d for the 2 digit format, no idea about the rest


Yes, thank DB. "%2d" give 2 digit format value, but not leading zero value if the value as 1 digit. So, I use "%02.f" to output all value with 2 digits.

I said, %.2d, not %2d

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Corroder
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PostPosted: Mon Jul 22, 2019 9:17 am    Post subject: Reply with quote

Dark Byte wrote:

I said, %.2d, not %2d


Ah, sorry, I don't see that dot before 2d. Yes, it works properly. Thanks, DB.

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