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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 | 293 views | 16 replies |

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

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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.

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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

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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");
?>
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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

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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");
?>
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Nathon Jones Web Design

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

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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?

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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?

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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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