Convert DateTime format to another DateTime format in Lua

print(os.date("%Y-%m-%d %H:%M:%S", "05/17/2017 05:17:00 PM"))

local date = '05/17/2017 05:17:00 PM'
local month, day , year, hours, minutes, seconds, amPm = date:match('^(%d%d)/(%d%d)/(%d%d%d%d) (%d%d):(% d%d):(%d%d) ([AP]M)$')
if not month then
 - Our entire match failed, and no captures were made
 error( 'could not parse date "'.. date ..'"')
end
if amPm =='PM' then
 hours = string.format('%2d', tonumber( hours) + 12)
end
local newDate = string .format(
'%s-%s-%s %s:%s:%s',
 year, month, day, hours, minutes, seconds
)
 print(newDate) - 2017-05-17 17:17:00

I have a problem where a date value is used to send to the application for some processing, which requires First format it into a different DateTime format.

Yes) I have

I start with the DateTime value in the following format:

> MM / DD / YYYY hh: mm: ss [AM / PM]

What I need

To use Lua in a certain way I need to convert it to DateTime format:

> YYYY-MM-DD hh:mm:ss

I have tried many different things, I have no luck; I will post what I have tried, I think this may be the most appropriate But maybe this is also off, and there is a simpler or simpler LUA method.

What did I try

print(os.date ("%Y-%m-%d %H:%M:%S","05/17/2017 05:17:00 PM"))

error

stdin:1: bad argument #2 to’date'(number expected, got string)

What am I assuming

< p>I think this means that the data type is a string instead of a datetime type, and it needs to be converted before os.date can use it. I also assume that I can pass the value of the datetime string to os.date, And it will output the same datetime value wisely, but I need to use the new format.

I have read various posts about the need to parse the datetime value part, and then use these parsed parts to Build the format you need. I hope someone can help guide me in the right direction with this task, so I don’t need to run through more trial and error and do more reading, but I will do it until I hear Come back or give up one day.

A brief review

>My value starts at: 05/17/2017 05:17:00 PM
>I want to send this value to 2017-05-17 17:17 :00

Using the newly formatted date and time, I can use it to process the application.

Other

Any level of Lua logic can solve this problem Will be explored, whether it is a custom function, different built-in functions to get the date time format, etc. I am open to all ideas, only print() and os.date (functions in my example) are shown , To keep it simple, but I am open to more complicated things.

Unlike other languages, Lua does not have a DateTime type; dates and times pass strings, numbers And strings and number tables are represented in Lua. The result of this is that there are many ways to use dates in Lua, and the best way depends on your situation.

You get from os. The error you get for date is because the second parameter should be a number in Unix time (the number of seconds since January 1, 1970). In order to work in your case, you need to use the os.time function to set the timestamp Converted to Unix time. However, os.time requires a table of numbers as input, so you need to do the actual date parsing and find the number from the date string.

But, in your case, you The input is a string, and the output you want is a similar string, so you only need to perform date parsing instead of using os.date and os.time to convert the results. This is what @tonypdmtr said in his answer What string.gsub does. The following is using string.match and string.format to do something very similar:

local date = '05/17/2017 05:17 :00 PM'
local month, day, year, hours, minutes, seconds, amPm = date:match('^(%d%d)/(%d%d)/(%d%d%d %d) (%d%d):(%d%d):(%d%d) ([AP]M)$')
if not month then
 - Our entire match failed , and no captures were made
 error('could not parse date "'.. date ..'"')
end
if amPm =='PM' then
 hours = string.format('%2d', tonum ber(hours) + 12)
end
local newDate = string.format(
'%s-%s-%s %s:%s:%s',
 year, month, day, hours, minutes, seconds
)
print(newDate) - 2017-05-17 17:17:00