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I got this new tutorial I just wrote. This version is in Python, I have this in many version. Currently I'm writing this small function into Ruby, C# and C. I have it in PHP, Javascript, Python, Perl. If you like it and want to see it in other format. Check out my blog @ http://kevinhng86.iblog.website. This simple script will demonstrate to you, how to convert a Date string into timestamp without needing to use anyone else library. It is not that big of a source code mostly because I write the tutorial inside it.

Code:

```
# dateToTimeStamp function work on yyyy-mm-dd
# This function features is to convert a date string in the format of yyyy-mm-dd pr yyyy/mm/dd to a unix timestamp.
# To read the entire tutorial please follow kevinhng86 blog @ http://kevinhng86.iblog.website .
# This tutorial is for Python and this source code is for Python.
# If you interest in learning time stamp converting in other programming language, look for it's on my website.
# Without this function you can use the built in Python function library.
# This function will work from the year 50 to 5500 due to memory allocation problem.
# Python built in library code:
# import time
# import datetime
# myDate = "2000-2-28"
# timestamp = time.mktime(datetime.datetime.strptime(myDate, "%Y-%m-%d").timetuple())
# print timestamp
# Python library draw back is it can't go below the year 1750
def dateToTimeStamp(value):
import re
out = 0 # This is the output value
datearray = [] # This array is use to split the input
if value is None or len(value) < 1: # No input value the script will die.
raise Exception("There is not an input value")
if re.match("^\d{4}-{1}\d{1,2}-{1}\d{1,2}$", value): # This check for 4 number for years plus a dash, 1 or 2 number for months plus a dash and 1 or 2 number for day plus a dash'''
datearray = value.split("-") # Split the number into an array.
elif re.match("^\d{4}\/{1}\d{1,2}\/{1}\d{1,2}$", value): # This check for 4 number for years plus a /, 1 or 2 number for months plus a / and 1 or 2 number for day plus a /
datearray = value.split("/") # Split the number into an array.
else:
raise Exception("Incorrect format") # If none of the above value match the script will exit with error
# Convert the string that store in the array into interger
inyyyy = int(datearray[0])
inmm = int(datearray[1])
indd = int(datearray[2])
# Check to see if the input year is leap year, century leap year or quadcentury leap year.
# If any of the condition is true the variable will store a one else it will store a 0.
isLeap = 1 if (inyyyy % 4) == 0 else 0
isCenturyLeap = 1 if (inyyyy % 100) == 0 else 0
isQuadCenturyLeap = 1 if (inyyyy % 400 ) == 0 else 0
# Can't have more than 12 months
if(inmm > 12):
raise Exception("Month cannot be greater than 12.")
# Can't have more than 31 days.
if(indd > 31):
raise Exception("Day cannot be great than 31.")
# Can't have more than 29 days in february
if (inmm == 2 and indd > 29):
raise Exception("February can't have more than 29 days")
# In April, June, September and November we only have 30 days.
if ( (inmm == 4 and indd > 30) or (inmm == 6 and indd > 30) or (inmm == 9 and indd > 30) or (inmm == 11 and indd > 30) ):
raise Exception("April, June, September and November can't have more than 30 days")
# Nothing can be less than 1
if (inmm < 1 or indd < 1 or inyyyy < 1):
raise Exception("Month date and years input can't be less than 1")
# Year can't be less than 50 due to memory allocation problem for storing one interger value
if (inyyyy < 50):
raise Exception("Memory doesn't allow us to convert year below 50")
# Year can't be less more than 5500 due to memory allocation problem for storing one interger value
if (inyyyy > 5500):
raise Exception("Memory doesn't allow us to convert year above 5500")
# If this is not a leap year than February can't have more than 28 days.
if (isLeap != 1 and inmm == 2 and indd > 28):
raise Exception("This is not a leap year, you can't have 29 days in Feburary")
# If this is a century leap year but is not a quad centuary leap year, we can't have more than 28 days.
if (inmm == 2 and isCenturyLeap == 1 and isQuadCenturyLeap != 1 and indd > 28 ):
raise Exception("This February can't have 29 days. Although it is a leap year, leap year is skipped on years that are divisible 100 years but will not skip if the year is divisible by 400")
# We set variable here in the event where the input value doesn't pass test. It won't consume much memory;
dateinmonth = [31,28,31,30,31,30,31,31,30,31,30,31] # This is how many day in each month. We count february at 28 because we will modify our equation to match the leap year value.
leapv = 0 # Variable for storing how many leap days occur since
aridx2 = inmm - 1 # This is the index we use to know which month we are on from the @dateinmonth array.
totaldate = 0 # Variable for storing the total date since 1970 1 1
yeardate = 0 # Variable for storing the year from
dateleftinyear = 0 # Variable for storing how many date left in the input value currentyear
dateleftinmonth = 0 # Variable for storing how date left in the input value current month
# So what is a timestamp. A timestamp is a number string that contain time value with the format of how many second since January 1, 1970.
# Timestamp actually measure in milisecond however for this tutorial we only use the second measurement.
# Negative timestamp is how far a time is below 1970. For instance any date in 1969 and below will generate a negative timestamp.
# Positive timestamp is how far a time is above 1970. For instance any date in 1970 and above will generate a positive timestamp. 1970 1 1 in this tutorial will generate a 0 timestamp.
# The first challenge we meet is leap year.
# Timestamp take into account leap years, century leaps and quad century leaps.
# What does leap years mean?
# We get the 29 day in February if the year is divisble by four, we dont get the 29 day if the year is divisible by 100 but not by 400.
#
# The method I prefer to use for calculating leap is if we are above 1970 we will use the closet below leap year, which is 1968 to calculate leap.
# This is the fact that we only need to round down our number after we minus year to 1968 and divide by four.
# If we use 1970 every time we hit a decimal .5 or .75 we have to round up.
#
# My formula for calculating year above 1970 is follow:
# Since we use leap to calculate to 1968 we will calculate to the 1st day of 1968 and then minus of two 365 days year from the equation.
# y = (input year value - 1968) * 365
# d = date from the begining of year = 365 - ( (total date in input month - input date) + ( how many date left in the year 1 month after input month);
# l = how many time leap year that divisble by 4 occur since 1968. Take input year - 1968 / 4 , the whole number here is what we want not the decimal. Since it is decimal it had not happen yet.
# c = how many time century leap occur since 1968. Input year - 1968 / 100, the whole number here is what we want not the decimal.
# q = how many time century leap occur since 1968. Input year - 1968 / 400, the whole number here is what we want not the decimal.
# totaldate = y - d + l - c + q ;
# timestamp = (totaldate * 86400) - 63158400 # 86400 is how many second in a day. and 63158400 is the second that contain within two years that is 365 days each.
# # Since we calculate to 1968 we have to give back this value.
#
# The formula for calculating the timestamp for if below 1970 will be put down below.
# For the purpose of this tutorial I divide the code into what happen if the year is above 1970 and what happen if it is below.
# Once you understand the concept it is possible to combine them and modify the equation.
#
if inyyyy >= 1970 :
leapv = int( (inyyyy - 1968)/4 ) # Get the whole interger for how many time leap year had occur since 1968.
centuryleapv = int( (inyyyy - 2000) / 100 ) # Get the whole interger for how many time century leap year had occur since 1968.
quadcenturyleapv = int( (inyyyy - 2000) / 400 ) # Get the whole interger for how many time quad century leap year had occur since 1968.
length = len(dateinmonth)
dateleftinmonth = dateinmonth[aridx2] - indd # Get how many date left in the current input month
for i in range(inmm, length):
dateleftinyear = dateleftinyear + dateinmonth[i] # Get how many date left in a the year since after the input month.
# Array index start at 0 therefore we do not need to add 1 to our indexer.
datefrombeginingofyear = 365 - (dateleftinmonth + dateleftinyear) # Get the date from the begining of year until input date by minus 365 to (dateleftinmonth + dateleftinyear)
# The real challenge we have for calculating leap is.
# On the year that it is a leap year, we do not have the extra day until we pass the 29 of february.
# Hence although in a leap year if it does not pass the 29 we can't add leap to our equation.
# However the automatic formula already took in account the leap year. Therefore those have to be minus off if it does not pass the 29 day of Feburary.
# Each if loop below is associate with a leap calculation.
#
if (isLeap == 1 and inmm < 3 ):
leapv = leapv - 1 if (inmm < 2) or (inmm == 2 and indd <= 29) else leapv
if (isCenturyLeap == 1 and inmm < 3 ):
centuryleapv = centuryleapv - 1 if (inmm < 2) or (inmm == 2 and indd <= 29) else centuryleapv
if (isQuadCenturyLeap == 1 and inmm < 3 ):
quadcenturyleapv = quadcenturyleapv - 1 if (inmm < 2) or (inmm == 2 and indd <= 29) else quadcenturyleapv
# After get all value we calculating. Assign the value to the out variable.
yeardate = (inyyyy - 1968) * 365
totaldate = yeardate + datefrombeginingofyear + leapv - centuryleapv + quadcenturyleapv
out = (totaldate * 86400) - 63158400
else:
# My formula for calculating year below 1970 is follow:
# The method I prefer to use for calculating leap is if we are below 1970 we will use the closet above leap year, which is 1972 to calculate leap.
# Since we use leap to calculate to we will calculate to the 1st day of 1968 and then minus of two 365 days year from the equation.
# y = (input year value - 1972 + 1 ) * 365 # We will have a negative value of how far we are away from 1972 but we have to add one here.
# # This is due to the fact that 1971 can be consider be one year away from 1972.
# # However it is not. We are still within the 365 day away but not really one year away.
# since now are calculating how far we are below.
# We simply just calculating how many days left in the year since our input value.
# This will get us how far we are away from the end of the year.
# d = date left in month = total day in input month - input date.
# e = date left in year = total left in the year after the input month.
#
# We want all the number below to be positive now.
# l = how many time leap year that divisble by 4 occur since 1968. 1972 - input year / 4 , the whole number here is what we want not the decimal. Since it is decimal it had not happen yet.
# c = how many time century leap occur since 1968. 1972 - input year / 100, the whole number here is what we want not the decimal.
# q = how many time century leap occur since 1968. 1972 - input year / 400, the whole number here is what we want not the decimal.
# The number now are in positive so that mean the further negative the further it is away and so forth.
# totaldate = y - d - e - l + c - q ;
# timestamp = ($totaldate * 86400) + 63158400 - 172800; # 86400 is how many second in a day. and 63158400 is the second that contain within two years that is 365 days each.
# # Since we calculate to 1972 we have to give back this value.
# # We have to minus two days to the equation which 172800 represent in second value.
# # In this formula if we are on the 31 day of december 1969 to 1970 1 1.
# # We will not have any day left in month or any day left in year.
# # Hence that is 0, but really we are on days away if we count by 0 hours.
# # Thus we have to minus of one day.
# # The same principle apply when we borrow the 2 years to 1972
#
#
leapv = int( (1972 - inyyyy)/4 )
centuryleapv = int( ( (2000 - inyyyy) / 100) )
quadcenturyleapv = int( (2000 - inyyyy) / 400 )
length = len(dateinmonth)
# The check leap here is difference since we going further down.
# Leap day will not be any further unless we pass march the first.
if ( isLeap == 1 and inmm > 2 ):
leapv = leapv - 1 if ( (inmm > 2) or (inmm == 3 and indd >= 1) ) else leapv
if (isCenturyLeap == 1 and inmm > 1 ):
centuryleapv = centuryleapv - 1 if ( (inmm > 2) or (inmm == 3 and indd >= 1) ) else centuryleapv
if (isQuadCenturyLeap == 1 and inmm > 1 ):
quadcenturyleapv = quadcenturyleapv - 1 if ( (inmm > 2) or (inmm == 3 and indd >= 1) ) else quadcenturyleapv
dateleftinmonth = dateinmonth[aridx2] - indd
for i in range(inmm,length):
dateleftinyear = dateleftinyear + dateinmonth[i]
yeardate = (inyyyy - 1972 + 1 ) * 365
totaldate = yeardate - dateleftinmonth - dateleftinyear - leapv + centuryleapv - quadcenturyleapv
out = (totaldate * 86400) + 63158400 - 172800
# Return output value.
# Most the the explanation already explain above.
# Stay tune for how to convert a timestamp back to date format without using a built in function.
return out
print dateToTimeStamp("1200-2-29")
```