Conversion Of An Id Number To The Persons Age?

[excel in]

I'm a 1st-time user, and I would like to thank you in advance for a formula which would help in future

My question relates to a column of ID no's with total of 13 numerals eg:5807085206087
I would like to convert this to the persons age eg: 57 yrs.

The first 6 numerals describe the birthdate,the first 2 numerals 58 (1958) the 2nd two numerals are the month (07) which is July and the (08) is the day.

I have a list of 3500 and I need to insert a formula to carry out this function in Excel 2013.

Could anybody assist with a formula

Many thanks and forever indebted to you for your assistance

 

[excel in]

[TABLE="width: 122"]
<tbody>[TR]
[TD="class: xl63, width: 122"]Howdy AliGW
The logic is in the example below using the formula.
id number 150101412208 gives this birthdate, 01-01-1915.[/TD]
[/TR]
</tbody>[/TABLE]

This person is only one year old
Regards and thanks

 

[jim may]

Seems like the question is: Do you have any centerians(persons over 100 years old) in your db?

 

[AliGW]

[TABLE="width: 122"]
<tbody>[TR]
[TD="class: xl63, width: 122"]Howdy AliGW
The logic is in the example below using the formula.
id number 150101412208 gives this birthdate, 01-01-1915.[/TD]
[/TR]
</tbody>[/TABLE]

This person is only one year old
Regards and thanks

That's not logical! There are still people alive today who are 110 or more, as Jim has suggested. You are going to have to do better than that.

 

[excel in]

Howdy AliGW
I am positive there are no people over 100 years in my Database. The oldest person in the Db is sixty six years and I know the child who was born in on the 1st January 2015.

 

[excel in]

Thank you

 

[excel in]

Howdy Jim may
I DO HAVE people less than 16 years of age. Could you possibly help with a revised formula to give the result 01/01/2015 for the id number 1501014122084.
Thank you for your input

 

[AliGW]

This will cater for anyone up to the age of 99:

=MID(A12,5,2)&"-"&MID(A12,3,2)&"- "&IF(LEFT(A12,2)<=TEXT(TODAY(),"yy"),"20"&LEFT(A12,2),"19"&LEFT(A12,2))

 

[excel in]

This will cater for anyone up to the age of 99:

=MID(A12,5,2)&"-"&MID(A12,3,2)&"- "&IF(LEFT(A12,2)<=TEXT(TODAY(),"yy"),"20"&LEFT(A12,2),"19"&LEFT(A12,2))

Thank you AliGW,
The formula works wonderful, except for id numbers beginning with zero. I have already tried formatting the id number.
An example below is the result I get. [TABLE="width: 411"]
<tbody>[TR]
[TD="class: xl64, width: 137"]e.g [/TD]
[TD="class: xl63, width: 274"]0601306114088 = 06-13-1960 which is incorrect,
The date of birth should be 30-01-2006.
[/TD]
[/TR]
</tbody>[/TABLE]
Thank you.

 

[AliGW]

So what distinguishes these new ID numbers? Are they the same length as other IDs or one digit longer? How is Excel supposed to know whether the leading zero is part of a date or an extra digit to be ignored? More to the point, how am I supposed to know?

Have you actually tried to adapt the formula I gave you yourself? What results did you get?

 

[excel in]

All id numbers are 13 digits in South africa, The leading digit zero is part of the date.
The e.g. 0601306114088. 06 = the year 2006, 01 = month = JANUARY, 30 =DAY=30th.
Hope you get the point. I have tried to adapt the formula. Excel is ignoring the zero.