How to persist and load NSDatePicker date values

Question

I am very new to MacOs development so apologies if this is a simplistic question:

I have an app that stores and loads data to an SQlite db.

A view has an NSDatePicker in graphic mode and the view can be either in 'add' or 'edit' mode ie adding or retrieving and updating data from SQlite

Another view has an NSTableView where one of the columns is the date that I want to load from SQlite into the column.

My question is what property of the datePicker should I persist - datePicker.stringValue or datepicker.dateValue?

If it is .stringValue can I simply load this when displaying the datePicker on the 'edit' view?

If it is .dateValue how do I get the value for the tableView? (use NSDateFormatter?)

SQlite does not have a date type so I am using a text field to store the date

How do you use the date? Which date and time elements are displayed by the picker? Which date format do you store in the database? — Willeke, Sep 21 '18 at 12:50
Adding a `NSDateFormatter` to the text field in the table view in the storyboard/XIB will make the object value of the text field a date. — Willeke, Sep 21 '18 at 12:51
I have no code written yet. The date will be used to display in a tableview column. I only want to show the date in the column not the time. In the SQlite database I am using the Text storage class as SQlite does not have a date storage class. I want the date to display in the table view as something like 3rd Jan 2018, but I need to be able to re-load the stored date from Sqlite into the datepicker — Colin W, Sep 21 '18 at 13:02

score 1 · Accepted Answer · answered Sep 21 '18 at 23:33

1

Converting dates back and forth to string is messy and somewhat fragile. I suggest saving it as a floating point value instead.

To convert a date to a Double use:

let interval = aDate.timeIntervalSinceReferenceDate

(Where aDate is a Date object, and interval is a Double, a.k.a a TimeInterval)

Conversely, to convert a floating point value back to a Date:

let aDate = Date(timeIntervalSinceReferenceDate: interval)

You should be able to save the resulting interval floating-point value to your SQL database. Use Double precision for the floating point value.

answered Sep 21 '18 at 23:33

Duncan C

Thanks very much that makes sense. Will try that out – Colin W Sep 24 '18 at 08:52
Have tried this and it works great. Thanks for the answer – Colin W Sep 25 '18 at 17:34
If my answer helped you you should accept it. That way other readers know which answer resolves somebody's problem. – Duncan C Sep 25 '18 at 18:54
Done that now. Didnt realise as new to Stack Overflow – Colin W Sep 26 '18 at 08:27
It's also considered good form to up-vote answers you find helpful, although that's entirely up to you. – Duncan C Sep 26 '18 at 10:26

1 Answers1