Converting a number todate

H

Hennie Neuhoff

In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits to
date of birth, without
any luck. The main problem being the year of birth forms the first 2 digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7 digits] .
Any help would be appreciated
HJN
 
R

Rick Rothstein \(MVP - VB\)

I am guessing no one in your country will ever live to be more than 100?
Here is a formula that seems to work for those less than 100 years old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick
 
B

Bob Phillips

=TEXT(A1,"yymmdd")

--
---
HTH

Bob


(there's no email, no snail mail, but somewhere should be gmail in my addy)
 
R

Rick Rothstein \(MVP - VB\)

You are going in the wrong direction... the OP has the number and wants to
produce a date from it... your formula assumes just the opposite (that the
OP has the date and wants to produce the number).

Rick


Bob Phillips said:
=TEXT(A1,"yymmdd")

--
---
HTH

Bob


(there's no email, no snail mail, but somewhere should be gmail in my
addy)



Hennie Neuhoff said:
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
 
R

Ron Coderre

Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
....Check: Fixed Width.......Click [Next]
....Insert a breakpoint after the 6th character (by clicking)...Click [Next]
....Select the 1st Col...Check: Date...YMD...Click [Next]
....Select the 2st Col...Check: Do not import
....Destinatioin: Select a cell off to the right of the 1st ID Number
....Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)
 
R

Rick Rothstein \(MVP - VB\)

I'm not sure where the 1900/2000 century switchover breakpoint is, but if
the cell contains this number...

2202103242344

as an example (the numbers from position 7 onward are immaterial), using
your Text To Column procedure will convert it to this...

2/10/2022

Based on what the OP said the first 6-digits represented, I would guess the
year should be 1922, not 2022.

Rick


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
...Check: Fixed Width.......Click [Next]
...Insert a breakpoint after the 6th character (by clicking)...Click
[Next]
...Select the 1st Col...Check: Date...YMD...Click [Next]
...Select the 2st Col...Check: Do not import
...Destinatioin: Select a cell off to the right of the 1st ID Number
...Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

message
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
 
R

Ron Coderre

Not sure if that's an issue or not.
We'll see, I guess.

--------------------------

Best Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)


Rick Rothstein (MVP - VB) said:
I'm not sure where the 1900/2000 century switchover breakpoint is, but if
the cell contains this number...

2202103242344

as an example (the numbers from position 7 onward are immaterial), using
your Text To Column procedure will convert it to this...

2/10/2022

Based on what the OP said the first 6-digits represented, I would guess
the year should be 1922, not 2022.

Rick


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
...Check: Fixed Width.......Click [Next]
...Insert a breakpoint after the 6th character (by clicking)...Click
[Next]
...Select the 1st Col...Check: Date...YMD...Click [Next]
...Select the 2st Col...Check: Do not import
...Destinatioin: Select a cell off to the right of the 1st ID Number
...Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

message
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
H

Hennie Neuhoff

Tanks. I've tried to do this with a macro. The objective being if the user
key in the ID no. to obtain the date of birth.
--
HJN


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
....Check: Fixed Width.......Click [Next]
....Insert a breakpoint after the 6th character (by clicking)...Click [Next]
....Select the 1st Col...Check: Date...YMD...Click [Next]
....Select the 2st Col...Check: Do not import
....Destinatioin: Select a cell off to the right of the 1st ID Number
....Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

Hennie Neuhoff said:
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
 
R

Rick Rothstein \(MVP - VB\)

In looking at this formula, I am pretty sure it does NOT work correctly, so
don't use it.

Rick


Rick Rothstein (MVP - VB) said:
I am guessing no one in your country will ever live to be more than 100?
Here is a formula that seems to work for those less than 100 years old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick


Hennie Neuhoff said:
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
 
R

Rick Rothstein \(MVP - VB\)

You should be able to use this function within your macro...

Function GetBirthday(IDnumber As String) As Date
Dim Yr As Long
Dim Mn As Long
Dim Dy As Long
Yr = Left(IDnumber, 2)
Mn = Mid(IDnumber, 3, 2)
Dy = Mid(IDnumber, 5, 2)
If Yr = Year(Now) Mod 2000 Then
Yr = 1900 - 100 * (DateSerial(Year(Now), Mn, Dy) <= Date) + Yr
Else
Yr = 1900 - 100 * (Yr < Year(Now) Mod 2000) + Yr
End If
GetBirthday = DateSerial(Yr, Mn, Dy)
End Function

You could use it something like this...

Sub TestMacro()
Dim Answer As String
Answer = InputBox("Enter ID number...")
If Answer Like "######*" Then
MsgBox "Birthday: " & GetBirthday(Answer)
End If
End Sub

Rick


Hennie Neuhoff said:
Tanks. I've tried to do this with a macro. The objective being if the user
key in the ID no. to obtain the date of birth.
--
HJN


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
....Check: Fixed Width.......Click [Next]
....Insert a breakpoint after the 6th character (by clicking)...Click
[Next]
....Select the 1st Col...Check: Date...YMD...Click [Next]
....Select the 2st Col...Check: Do not import
....Destinatioin: Select a cell off to the right of the 1st ID Number
....Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

message
In our country all citizens have a unique [13digit] identity number,
the
first 6 digits being date of bitrh. I've tried to convert these 6
digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
R

Rick Rothstein \(MVP - VB\)

Here is a corrected formula that will do what I intended my first (flawed)
formula to do...

=DATE(IF(LEFT(A1,2)=RIGHT(YEAR(NOW()),2),1900+100*(DATE(YEAR(NOW()),MID(A1,3,2),MID(A1,5,2))<=TODAY())+LEFT(A1,2),1900+100*(LEFT(A1,2)<RIGHT(YEAR(NOW()),2))+LEFT(A1,2)),MID(A1,3,2),MID(A1,5,2))

Rick


Rick Rothstein (MVP - VB) said:
I am guessing no one in your country will ever live to be more than 100?
Here is a formula that seems to work for those less than 100 years old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick


Hennie Neuhoff said:
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
 
R

Ron Coderre

Maybe this:

With A1 formatted as TEXT

B1:
=--(TEXT(MID(A1,3,4),"00\/00\/")&IF(--(TEXT(MID(A1,3,4),"00\/00\/")&
"20"&LEFT(A1,2))>TODAY(),19,20)&LEFT(A1,2))

Does that help?
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
...Check: Fixed Width.......Click [Next]
...Insert a breakpoint after the 6th character (by clicking)...Click
[Next]
...Select the 1st Col...Check: Date...YMD...Click [Next]
...Select the 2st Col...Check: Do not import
...Destinatioin: Select a cell off to the right of the 1st ID Number
...Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

message
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
 
R

Rick Rothstein \(MVP - VB\)

Very nice! Your formula seems to generate the same values mine does, but
yours is shorter and uses a lot less function calls.

Rick


Ron Coderre said:
Maybe this:

With A1 formatted as TEXT

B1:
=--(TEXT(MID(A1,3,4),"00\/00\/")&IF(--(TEXT(MID(A1,3,4),"00\/00\/")&
"20"&LEFT(A1,2))>TODAY(),19,20)&LEFT(A1,2))

Does that help?
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
...Check: Fixed Width.......Click [Next]
...Insert a breakpoint after the 6th character (by clicking)...Click
[Next]
...Select the 1st Col...Check: Date...YMD...Click [Next]
...Select the 2st Col...Check: Do not import
...Destinatioin: Select a cell off to the right of the 1st ID Number
...Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

message
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
R

Ron Rosenfeld

In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits to
date of birth, without
any luck. The main problem being the year of birth forms the first 2 digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7 digits] .
Any help would be appreciated
HJN
Click to expand...

For a formula, perhaps:

=DATE(INT(A1/10^11)+1900+100*(INT(A1/10^11)<YEAR(TODAY())/100),
MOD(INT(A1/10^9),100),MOD(INT(A1/10^7),100))


--ron
 
R

Ron Coderre

Thanks,Rick....Not being an Excel "date formula guru", these are fun to play
with.

I worked on it a little more and came up with this:
=LOOKUP(TODAY(),INDEX(--(TEXT(MID(A1,3,4),"00\/00\/")&{"19","20"}&LEFT(A1,2)),0))

--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)


Rick Rothstein (MVP - VB) said:
Very nice! Your formula seems to generate the same values mine does, but
yours is shorter and uses a lot less function calls.

Rick


Ron Coderre said:
Maybe this:

With A1 formatted as TEXT

B1:
=--(TEXT(MID(A1,3,4),"00\/00\/")&IF(--(TEXT(MID(A1,3,4),"00\/00\/")&
"20"&LEFT(A1,2))>TODAY(),19,20)&LEFT(A1,2))

Does that help?
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)


Ron Coderre said:
Try this to create a list of the dates off to the right of the original
values:

Select the single column range of ID Numbers

From the Excel Main Menu:
<data><text-to-columns>
...Check: Fixed Width.......Click [Next]
...Insert a breakpoint after the 6th character (by clicking)...Click
[Next]
...Select the 1st Col...Check: Date...YMD...Click [Next]
...Select the 2st Col...Check: Do not import
...Destinatioin: Select a cell off to the right of the 1st ID Number
...Click [Finish]


Is that something you can work with?
Post back if you have more questions.
--------------------------

Regards,

Ron
Microsoft MVP (Excel)
(XL2003, Win XP)

message
In our country all citizens have a unique [13digit] identity number,
the
first 6 digits being date of bitrh. I've tried to convert these 6
digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
G

Gord Dibben

I just ran 520210987654 through Data>Text to Column>Fixed Width.

Select first 6 numbers and Column Data Format>YMD.

Select other column and "Skip" then Finish.

February 10, 1952 was the result in Column A

After formatting of course.


Gord Dibben MS Excel MVP


Here is a corrected formula that will do what I intended my first (flawed)
formula to do...

=DATE(IF(LEFT(A1,2)=RIGHT(YEAR(NOW()),2),1900+100*(DATE(YEAR(NOW()),MID(A1,3,2),MID(A1,5,2))<=TODAY())+LEFT(A1,2),1900+100*(LEFT(A1,2)<RIGHT(YEAR(NOW()),2))+LEFT(A1,2)),MID(A1,3,2),MID(A1,5,2))

Rick


Rick Rothstein (MVP - VB) said:
I am guessing no one in your country will ever live to be more than 100?
Here is a formula that seems to work for those less than 100 years old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick


Hennie Neuhoff said:
In our country all citizens have a unique [13digit] identity number, the
first 6 digits being date of bitrh. I've tried to convert these 6 digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
R

Rick Rothstein \(MVP - VB\)

Thanks,Rick....Not being an Excel "date formula guru", these are fun
to play with.
Click to expand...

Yes, struggling to find shorter and shorter formula solutions is definitely
a fun process.
I worked on it a little more and came up with this:
=LOOKUP(TODAY(),INDEX(--(TEXT(MID(A1,3,4),"00\/00\/")&{"19","20"}&LEFT(A1,2)),0))
Click to expand...

A nice, thinking-outside-of-the-box solution. Here is yet another formula
that turns out to be 3 characters shorter than your posted (still using 6
function calls though)...

=DATEVALUE((20-(LEFT(A1,6)>TEXT(NOW(),"yymmdd")))&TEXT(LEFT(A1,6),"00-00-00"))

However, if you replace your TODAY function call with NOW (I don't think
that affect the functionality of formula any), then my formula becomes only
1 character shorter.

Rick
 
R

Rick Rothstein \(MVP - VB\)

Ron posted that process a little later on in this thread and I responded
with what I thought was an incorrect result from it (given the OP's stated
meaning of the first 6 digits). The example I gave was the first six digits
being 220210 and the fact that Text To Column yielded a date of 2/10/2022
instead of 2/10/1922 which is what I presume the correct result should be.

Rick


Gord Dibben said:
I just ran 520210987654 through Data>Text to Column>Fixed Width.

Select first 6 numbers and Column Data Format>YMD.

Select other column and "Skip" then Finish.

February 10, 1952 was the result in Column A

After formatting of course.


Gord Dibben MS Excel MVP


Here is a corrected formula that will do what I intended my first (flawed)
formula to do...

=DATE(IF(LEFT(A1,2)=RIGHT(YEAR(NOW()),2),1900+100*(DATE(YEAR(NOW()),MID(A1,3,2),MID(A1,5,2))<=TODAY())+LEFT(A1,2),1900+100*(LEFT(A1,2)<RIGHT(YEAR(NOW()),2))+LEFT(A1,2)),MID(A1,3,2),MID(A1,5,2))

Rick


Rick Rothstein (MVP - VB) said:
I am guessing no one in your country will ever live to be more than 100?
Here is a formula that seems to work for those less than 100 years old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick


message In our country all citizens have a unique [13digit] identity number,
the
first 6 digits being date of bitrh. I've tried to convert these 6
digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
G

Gord Dibben

Missed that part of the thread.

Change the Regional Settings in Windows if the years will be prior to current
range of 1939 to 2039 or whaterver.

Yeah, I know..........that is a pain in the butt.


Gord

Ron posted that process a little later on in this thread and I responded
with what I thought was an incorrect result from it (given the OP's stated
meaning of the first 6 digits). The example I gave was the first six digits
being 220210 and the fact that Text To Column yielded a date of 2/10/2022
instead of 2/10/1922 which is what I presume the correct result should be.

Rick


Gord Dibben said:
I just ran 520210987654 through Data>Text to Column>Fixed Width.

Select first 6 numbers and Column Data Format>YMD.

Select other column and "Skip" then Finish.

February 10, 1952 was the result in Column A

After formatting of course.


Gord Dibben MS Excel MVP


Here is a corrected formula that will do what I intended my first (flawed)
formula to do...

=DATE(IF(LEFT(A1,2)=RIGHT(YEAR(NOW()),2),1900+100*(DATE(YEAR(NOW()),MID(A1,3,2),MID(A1,5,2))<=TODAY())+LEFT(A1,2),1900+100*(LEFT(A1,2)<RIGHT(YEAR(NOW()),2))+LEFT(A1,2)),MID(A1,3,2),MID(A1,5,2))

Rick


message I am guessing no one in your country will ever live to be more than 100?
Here is a formula that seems to work for those less than 100 years old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick


message In our country all citizens have a unique [13digit] identity number,
the
first 6 digits being date of bitrh. I've tried to convert these 6
digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
Click to expand...
Click to expand...
Click to expand...
 
R

Rick Rothstein \(MVP - VB\)

Not to mention making that change would affect every other application that
uses 2-digit dates on the user's system from then on out. Or were you
proposing changing it temporarily? If so, I got the impression that the OP
needed to do this conversion to birthdates more than just one time (he
mentioned the user keying in the code to get the date via a macro).

Rick


Gord Dibben said:
Missed that part of the thread.

Change the Regional Settings in Windows if the years will be prior to
current
range of 1939 to 2039 or whaterver.

Yeah, I know..........that is a pain in the butt.


Gord

Ron posted that process a little later on in this thread and I responded
with what I thought was an incorrect result from it (given the OP's stated
meaning of the first 6 digits). The example I gave was the first six
digits
being 220210 and the fact that Text To Column yielded a date of 2/10/2022
instead of 2/10/1922 which is what I presume the correct result should be.

Rick


Gord Dibben said:
I just ran 520210987654 through Data>Text to Column>Fixed Width.

Select first 6 numbers and Column Data Format>YMD.

Select other column and "Skip" then Finish.

February 10, 1952 was the result in Column A

After formatting of course.


Gord Dibben MS Excel MVP


On Mon, 11 Feb 2008 20:17:32 -0500, "Rick Rothstein \(MVP - VB\)"

Here is a corrected formula that will do what I intended my first
(flawed)
formula to do...

=DATE(IF(LEFT(A1,2)=RIGHT(YEAR(NOW()),2),1900+100*(DATE(YEAR(NOW()),MID(A1,3,2),MID(A1,5,2))<=TODAY())+LEFT(A1,2),1900+100*(LEFT(A1,2)<RIGHT(YEAR(NOW()),2))+LEFT(A1,2)),MID(A1,3,2),MID(A1,5,2))

Rick


in
message I am guessing no one in your country will ever live to be more than
100?
Here is a formula that seems to work for those less than 100 years
old...

=DATE(1900+100*(IF(LEFT(A1,2)<=RIGHT(YEAR(NOW()),2),IF(--MID(A1,3,2)<=MONTH(NOW()),IF(--MID(A1,5,2)<=DAY(NOW()),1,0),0),0))+LEFT(A1,2),MID(A1,3,2),MID(A1,5,2))

Rick


message In our country all citizens have a unique [13digit] identity number,
the
first 6 digits being date of bitrh. I've tried to convert these 6
digits
to
date of birth, without
any luck. The main problem being the year of birth forms the first 2
digits,
ie a person born on 10 February 1952 IDnumber would be 520210[plus 7
digits] .
Any help would be appreciated
HJN
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