# Calculating someone's age by comparing DOB in a recordset with current date and display it in years and months?

Thread began 6/01/2016 5:32 pm by Nathon Jones Web Design | Last modified 7/05/2016 11:15 am by Ray Borduin | 475 views | 16 replies |

## 6/01/2016 5:32 pm  |  #1 Nathon Jones Web Design

### Calculating someone's age by comparing DOB in a recordset with current date and display it in years and months?

Is this complicated? I've found some functions online but none of them use a recordset value, they assume a static value.

This, for example:

`<?php  //date in mm/dd/yyyy format; or it can be in other formats as well  \$birthDate = "12/17/1983";  //explode the date to get month, day and year  \$birthDate = explode("/", \$birthDate);  //get age from date or birthdate  \$age = (date("md", date("U", mktime(0, 0, 0, \$birthDate[0], \$birthDate[1], \$birthDate[2]))) > date("md")    ? ((date("Y") - \$birthDate[2]) - 1)    : (date("Y") - \$birthDate[2]));  echo "Age is:" . \$age;?>`

I changed to this...

`<?php  //date in mm/dd/yyyy format; or it can be in other formats as well  \$birthDate = \$rsPROFILE->getColumnVal("dogsdob");  //explode the date to get month, day and year  \$birthDate = explode("/", \$birthDate);  //get age from date or birthdate  \$age = (date("md", date("U", mktime(0, 0, 0, \$birthDate[0], \$birthDate[1], \$birthDate[2]))) > date("md")    ? ((date("Y") - \$birthDate[2]) - 1)    : (date("Y") - \$birthDate[2]));  echo \$age;?>`

..but all it does is display "2016".

Could anyone point me in the direction of a decent function for this that can handle UK date formats?
Thank you.
NJ

## 6/02/2016 12:38 am  |  #2 Ray BorduinWebAssist

<?php
\$birthDate = strtotime("12/17/1983");
\$ageInSeconds = time() - \$birthDate;
\$age = intval(\$ageInSeconds/(60*60*24*365));
?>

There is a small issue with leap years where it could be off by a few days depending on how many leap years have passed.

## 6/02/2016 10:21 am  |  #3 Nathon Jones Web Design

A few days is neither here nor there as it's for dogs.

I changed your wee function (it is called a function right?) to this:

<?php
\$birthDate = strtotime(\$rsPROFILE->getColumnVal("dogsdob"));
\$ageInSeconds = time() - \$birthDate;
\$age = intval(\$ageInSeconds/(60*60*24*365));
?>

...but the result is blank?
Is it possible to output the result as:
12 years 5 months

...obviously hiding years or months if either is 0.
Jeez...maybe more to this than I thought.

Thanks for the help.
NJ

## 6/02/2016 2:52 pm  |  #4 Ray BorduinWebAssist

php:
``````<?php
\$birthDate = strtotime(\$rsPROFILE->getColumnVal("dogsdob"));
\$ageInSeconds = time() - \$birthDate;
\$age = intval(\$ageInSeconds/(60*60*24*365));
\$months = intval(((\$ageInSeconds/(60*60*24)) % 365)/12);
echo(strval(\$age) . " years " . \$months . " months");
?>``````

## 6/02/2016 7:09 pm  |  #5 Nathon Jones Web Design

I don't think that's dealing with UK date formats. See link below.
This particular record's dob, in the MySQL table, is 20/08/2011.
Thank you Ray.
NJ

## 6/02/2016 7:35 pm  |  #6 Ray BorduinWebAssist

Is it a varchar column in the database? php will assume european date format if you use dashes instead of slashes I think, so you may be able to use:

php:
``````<?php
\$birthDate = strtotime(str_replace("/","-",\$rsPROFILE->getColumnVal("dogsdob")));
\$ageInSeconds = time() - \$birthDate;
\$age = intval(\$ageInSeconds/(60*60*24*365));
\$months = intval(((\$ageInSeconds/(60*60*24)) % 365)/12);
echo(strval(\$age) . " years " . \$months . " months");
?>``````

## 6/02/2016 7:38 pm  |  #7 Nathon Jones Web Design

No, it's DATETIME.
There is no change with that revised code.

## 6/03/2016 12:49 am  |  #8 Ray BorduinWebAssist

Use this code:

php:
``````<?php
\$birthDate = strtotime(\$rsPROFILE->getColumnVal("dogsdob"));
\$ageInSeconds = time() - \$birthDate;
\$age = intval(\$ageInSeconds/(60*60*24*365));
\$months = intval(((\$ageInSeconds/(60*60*24)) % 365)/12);
echo("Birthdate: " . \$rsPROFILE->getColumnVal("dogsdob") . " age: " . strval(\$age) . " years " . \$months . " months");
?>``````

What does it display?

## 6/03/2016 3:04 am  |  #9 Nathon Jones Web Design

THAT code has a syntax error at line 6 I'm afraid.
I'm not clear why you've re-added the database value in line 6 either?

## 6/03/2016 3:08 am  |  #10 Ray BorduinWebAssist

I've fixed the syntax error. I've written it to the page to see what the value is coming from the database so I can adjust the code if necessary.

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