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[PHP/SQL] Calculate Age

17 posts in this topic

Posted

I have a mySQL database and a PHP frontend. The date of birth is stored within the database and im looking for some code that calculates the age preferably as a PHP

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Posted

Where dob is the column which holds the date of birth. :)


SELECT DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(dob, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(dob, '00-%m-%d')) AS age

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Posted

Never done this but could be something along:


age = current_year - birth_year;


if(current_date < birth_date && current_month < birth_month){

    age = age - 1;

}

Assuming you store months with their numerical value which is the better way IMHO.

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Posted

The PHP website is reading straight off the table called members. I havent created any views or anything. Im also using phpMyAdmin to create all the database.

Where would I type in that statement?

SELECT DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(dob, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(dob, '00-%m-%d')) AS age

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Posted

You would execute it where required; it's no different to any of the other queries you're executing.


SELECT

    id

  , DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(dob, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(dob, '00-%m-%d')) AS age

  , name

FROM

  member

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Posted

Are you not encrypting those DOBs?

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Posted

You would execute it where required; it's no different to any of the other queries you're executing.


SELECT

    id

  , DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(dob, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(dob, '00-%m-%d')) AS age

  , name

FROM

  member

There are no queries. which means im going somethign wrong here :s

Ok... Let me explain this situation a little better.

The database has information on members and their date of birth is stored. I have created a form using PHPRunner which creates the form from the members table. Since age cannot be stored into the database. The software has the ability to input functions or PHP code snippets.

So i was thinking, rather can creating queries, i would use php code instead to calculate the age. The php website can add and view members from the database.

hope this helps

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Posted

you can just use DATEDIFF, and then round the end figure down after dividing the number of days in a year

i.e. my birthday is 11 oct 1984

select datediff('11-oct-1984',now())/365

that query results in 26, which is correct.. i am not a MySQL or PHP Dev so please test this first, it works in MSSQL anyway :p

There are no queries. which means im going somethign wrong here :s

well, how are you going to get the data out of the db? magic? :blink:

you will need to query the db in order to get the DOB, so, why not in your query convert it to the age rather than get php to do it for you on the page?

if you want to get the DOB from the db, and then convert it on the page via PHP.. try this function: http://www.bradino.com/php/calculate-age/ ;)

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Posted

duplicate :)

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Posted

I searched and found this: http://php.net/manual/en/datetime.diff.php

EDIT: Guy above me was faster XD.

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Posted

I see.


function getAge($birthday){

  list($day, $month, $year) = explode('/', $birthday);

  $age        = date('Y') - $year;

  $month_diff = date('m') - $month;

  $day_diff   = date('d') - $day;

  if($day_diff < 0 || $month_diff < 0){

    $age--;

  }

  return $age;

}

I was going to suggest DateTime:diff but it is 5.3 dependent.*

*date_diff is an alias for DateTime::diff

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Posted

@Anthony:

Shouldn't that be:


if($day_diff < 0 && $month_diff < 0)

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Posted

Thanks guys, got it working. I used the following query to calculate the code


SELECT *, DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(aDoB, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(aDoB, '00-%m-%d')) AS age FROM members

And then I saved it as a View. I then created the PHP table using the view and now the age automatically update.

TYVM

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Posted

Thanks guys, got it working. I used the following query to calculate the code


SELECT *, DATE_FORMAT(NOW(), '%Y') - DATE_FORMAT(aDoB, '%Y') - (DATE_FORMAT(NOW(), '00-%m-%d') < DATE_FORMAT(aDoB, '00-%m-%d')) AS age FROM members

And then I saved it as a View. I then created the PHP table using the view and now the age automatically update.

TYVM

looks a bit inefficient, not sure though, would have to test it out

glad it works :)

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Posted

you can just use DATEDIFF, and then round the end figure down after dividing the number of days in a year

i.e. my birthday is 11 oct 1984

select datediff('11-oct-1984',now())/365

that query results in 26, which is correct.. i am not a MySQL or PHP Dev so please test this first, it works in MSSQL anyway :p

Leap years will throw that off so it is inaccurate.

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Posted

Leap years will throw that off so it is inaccurate.

hmm.. correct :(

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Posted

hmm.. correct :(

You could always determine how many leap years have passed between the date in question and the current date and then subtract that -- but that just seems... silly ;) (and actually even then it might be wrong...)

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