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User Profile: Birthday VS Anniversary

database,user,user-profile,date-of-birth,fullcontact

Usually the "Birthday Represent the date of birth while the "Anniversary" represents the day of marriage.

Difference between birthday formulas in SQL server programming

sql-server,formula,date-of-birth

It looks like the original code is not taking today into account, it always starts tomorrow, which may be one reason why your new code is finding more people. However, consider your logic: DATEPART(DY,dateOfBirth) - DATEPART(DY, CURRENT_TIMESTAMP) > 0 According to this, if my date of birth was January 1st...

How to retrieve current date and day_of_month from date in hsqldb for listing upcoming birthdays

java,date,hsqldb,dayofweek,date-of-birth

The first use of INTERVAL is incorrect syntax. You can modify it like this: SELECT * FROM contacts WHERE DATE_ADD(dob, (CURDATE() - dob) YEAR) BETWEEN CURDATE() AND DATE_ADD(CURDATE(), INTERVAL 7 DAY); ...

Calculation Of Days Difference Between BirthDay And Current Day

php,date-of-birth

The least verbose way to do this is with the DateTime classes and specifically with the DateTime::diff method. Note you should ensure the times are midnight for each date, then test if the birthday this year has already occurred. <?php $birthday = "1980-06-24"; // get date of birthday this calendar...

Validating Date of birth signup calendar

javascript,jquery,html,drop-down-menu,date-of-birth

If you want to do this manually.. This code should suffice (I'm pretty sure the pattern supports dd/mm/yyyy). function checkDateTime(newDate) { var pattern = '^(?:(?:31(\/|-|\.)(?:0?[13578]|1[02]))\1|(?:(?:29|30)(\/|-|\.)(?:0?[1,3-9]|1[0-2])\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:29(\/|-|\.)0?2\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:0?[1-9]|1\d|2[0-8])(\/|-|\.)(?:(?:0?[1-9])|(?:1[0-2]))\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$'; var regex = new RegEx(pattern); if (typeof...

AlarmManager - How to set a Notification to appear annually

android,notifications,alarmmanager,date-of-birth

You could replace 'HERE' with a method that determines if the following February from today is in a leap year, and then returns the value 365 or 366 days (in the form of milliseconds mind you) based on those checks. private long millisUntilNextYear(){ //Set days in a year for Leap...

change a column from birth date to age in r

r,date-of-birth

From the comments of this blog entry, I found the age_calc function in the eeptools package. It takes care of edge cases (leap years, etc.), checks inputs and looks quite robust. > library(eeptools) > x <- as.Date(c("2011-01-01", "1996-02-29")) > age_calc(x) # default is age in months [1] 46.73333 224.83118 >...

calculating age and turning it into a variable

javascript,validation,date-of-birth

The reason you see many examples using a "return" goes along with the advantages of structuring your code into functions. This would be a good next step for you in learning to program. Functions will allow you to create chunks of code that are reusable and also easier to read...

How do I add a date of birth UIDatePicker to UITextField?

ios,objective-c,uitextfield,date-of-birth

1) in the .h file of your view controller make sure you assign textfield delegate: @interface YourViewController : UIViewController <UITextFieldDelegate> and also have an IBOutlet of the birthday Textfield 2) declare a date picker as a class variable to make it accessible from all different methods in class. in the...

Filter people from Django model using birth date

python,django,date,filter,date-of-birth

You'll have to check against a datetime.date that is the maximum allowed birth date: anyone born after that day will be younger than the min age: from datetime import date min_age = 24 max_date = date.today() try: max_date = max_date.replace(year=max_date.year - min_age) except ValueError: # 29th of february and not...

Finding age with date of birth and grouping in terms of age in R

r,grouping,date-of-birth

Try indx <- round(as.numeric(difftime(Sys.Date(), df$DOB, unit='weeks'))/52.25) df$grp <- cut(indx, breaks=c(0,13,18,25,Inf), labels=c('kid', 'Teen', 'Young Adult', 'Old') ) ...