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T-SQL: calculating person age

Posted on 2013-11-01
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Last Modified: 2013-11-19
One table column has Date of Birth.
I have to calculate person's age in  years, for example 1.23 ,  8.35 , 37.36 etc. in another column.
Any thoughts on this?
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Question by:quasar_ee
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Simone B earned 250 total points
ID: 39617946
DATEDIFF returns an integer, so you have to cast or convert to a decimal. If you use 365.25 days per year instead of 365, that will cover for the leap years. Then to round the whole thing to 2 decimal places, cast again:

SELECT CAST(CAST(DATEDIFF(d,YourDateField,GETDATE()) AS DECIMAL(10,2))/365.00 AS DECIMAL(10,2))
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Expert Comment

by:Patrick Matthews
ID: 39619158
Buttercup1,

I like it, but I see a couple of problems, such as:
In either formula, having two decimal places is not enough precision, and
In the 365.25 formula, you have a risk that you will understate the true age in some edge cases

Illustrating the first, consider this example:

DECLARE @dob datetime = '2012-02-28',
    @today datetime = '2013-02-27'
    
SELECT CAST(CAST(DATEDIFF(d, @dob, @today) AS DECIMAL(10, 2)) / 365.00 AS DECIMAL(10, 2)) AS _365,
    CAST(CAST(DATEDIFF(d, @dob, @today) AS DECIMAL(10, 2)) / 365.25 AS DECIMAL(10, 2)) AS _36525

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This returns 1.00 and 1.00, respectively, yet most people would agree that this person is not yet a year old.  Using only two digits for the decimal portion is not precise enough.  Using decimal(10, 4) instead would remedy that.

Now consider how using 365.25 can understate the age in some edge cases.  (I am adopting the more precise decimal(10, 4) here.)

DECLARE @dob datetime = '2013-02-28',
    @today datetime = '2014-02-28'
    
SELECT CAST(CAST(DATEDIFF(d, @dob, @today) AS DECIMAL(10, 4)) / 365.00 AS DECIMAL(10, 4)) AS _365,
    CAST(CAST(DATEDIFF(d, @dob, @today) AS DECIMAL(10, 4)) / 365.25 AS DECIMAL(10, 4)) AS _36525

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The "365" formula returns 1.0000, and the "365.25" formula returns 0.9993, but everyone would agree that such a person should be counted as one year old.

:)

Patrick
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LVL 92

Assisted Solution

by:Patrick Matthews
Patrick Matthews earned 250 total points
ID: 39619233
Inelegant, and there may well be a better way, but this does address the points I made above:

SELECT 
    DOB, GETDATE(),
    DATEADD(year, DATEDIFF(year, DOB, GETDATE()), DOB),
    DATEADD(year, DATEDIFF(year, DOB, GETDATE()) - 1, DOB),
    CASE 
    WHEN CONVERT(date, DATEADD(year, DATEDIFF(year, DOB, GETDATE()), DOB)) > CONVERT(date, GETDATE()) THEN
        CONVERT(decimal(10, 4), DATEDIFF(year, DOB, GETDATE()) - 1) + 
        CONVERT(decimal(10, 4), DATEDIFF(day, DATEADD(year, DATEDIFF(year, DOB, GETDATE()) - 1, DOB), GETDATE())) / 
        CONVERT(decimal(10, 4), DATEDIFF(day, DATEADD(year, DATEDIFF(year, DOB, GETDATE()) - 1, DOB), DATEADD(year, DATEDIFF(year, DOB, GETDATE()), DOB)))
    WHEN CONVERT(date, DATEADD(year, DATEDIFF(year, DOB, GETDATE()), DOB)) < CONVERT(date, GETDATE()) THEN
        CONVERT(decimal(10, 4), DATEDIFF(year, DOB, GETDATE())) +
        CONVERT(decimal(10, 4), DATEDIFF(day, DATEADD(year, DATEDIFF(year, DOB, GETDATE()), DOB), GETDATE())) / 
        CONVERT(decimal(10, 4), DATEDIFF(day, DATEADD(year, DATEDIFF(year, DOB, GETDATE()), DOB), DATEADD(year, DATEDIFF(year, DOB, GETDATE()) + 1, DOB)))
    ELSE
        CONVERT(decimal(10, 4), DATEDIFF(year, DOB, GETDATE()))
    END AS Age
FROM SomeTable

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