I need a genius for this one!

[ameshkin]

What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

How do I do this. My brain hurts enough already from writing the
function above. Can anyone help?

[Surfer!]

In message <11**********************@h48g2000cwc.googlegroups .com>,
ameshkin <am**********@gmail.comwrites

>What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

How do I do this. My brain hurts enough already from writing the
function above. Can anyone help?

Do it in the SQL you use to select from the database. Of course this
assumes you have used a date-type column to hold the date of birth!

For MySQL see:
http://dev.mysql.com/doc/refman/5.0/...functions.html

Check the YEAR function.

--
Surfer!
Email to: ramwater at uk2 dot net

[Mitul]

ameshkin wrote:

What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

How do I do this. My brain hurts enough already from writing the
function above. Can anyone help?

Hi ameshkin.

Here is the approx solution to ur answer. You can do things by Query.
Here is the query for that You needs to just modify little by PHP and
Pass it on to the My-SQL u will get poper age.

SELECT name, birth, CURRENT_DATE,
-(YEAR(CURRENT_DATE)-YEAR(birth))
-- (RIGHT(CURRENT_DATE,5)<RIGHT(birth,5))
-AS age
-FROM pet;
/***** There is nothing impossible coz Impossible itself means I M
Possible. :)
Regards,
Mitul Patel

[Oli Filth]

ameshkin wrote:

What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

I assume from the above that you are storing birthdays in the database
as strings. This is more trouble than it's worth. Most databases have
a date type, and functions which allow easy comparisons, etc.

MySQL, for instance, has loads:
http://dev.mysql.com/doc/refman/5.1/...functions.html
--
Oli

[ameshkin]

Yes, thats one problem. For other reasons, teh date is in the database
as a string. When I get home, i will changbe the structure around and
see if i can just use a simple sql query to get my results.

Thanks for the help
Oli Filth wrote:

ameshkin wrote:

What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

I assume from the above that you are storing birthdays in the database
as strings. This is more trouble than it's worth. Most databases have
a date type, and functions which allow easy comparisons, etc.

MySQL, for instance, has loads:
http://dev.mysql.com/doc/refman/5.1/...functions.html
--
Oli

[IchBin]

ameshkin wrote:

What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

How do I do this. My brain hurts enough already from writing the
function above. Can anyone help?

Friend!, if you have to cross post make sure you have all of the
newsgroups in to TO: statement before you send it. It stops you from
wasting other peoples time not knowing someone else had resolved your
problem already in another group!

First off, you should keep the date in the database the default format
'yyyy-MM-DD'.

I am not handling leap years but gives you a start. You may want to
build a procedure to do all of this and include leap year:

$minAge = 24;
$maxAge = 54;

SELECT
LASTNAME,
FIRSTNAME,
ROUND(DATEDIFF(CURDATE(),birthday)/365) AS 'numberOfYears',
birthday
FROM table
WHERE
ROUND(DATEDIFF(CURDATE(),birthday)/365) <= $maxAge AND
ROUND(DATEDIFF(CURDATE(),birthday)/365) >= $minAge;

--
Thanks in Advance...
IchBin, Pocono Lake, Pa, USA http://weconsultants.phpnet.us
__________________________________________________ ________________________

'If there is one, Knowledge is the "Fountain of Youth"'
-William E. Taylor, Regular Guy (1952-)

[IchBin]

ameshkin wrote:

What I am trying to do, is...

Find users in a database between two ages.

$minage
$maxage

The birthdays of the users are in the database in this format
01-30-1980

Now this function takes a birthday in this format 1980-01-30 and
returns the age

function birthday ($birthday)
{
list($year,$month,$day) = explode("-",$birthday);
$year_diff = date("Y") - $year;
$month_diff = date("m") - $month;
$day_diff = date("d") - $day;
if ($month_diff < 0) $year_diff--;
elseif (($month_diff==0) && ($day_diff < 0)) $year_diff--;
return $year_diff;
}

But what I need to do, is somehow only show the results from the
database where the birthday is in between two different ages.

How do I do this. My brain hurts enough already from writing the
function above. Can anyone help?

Friend!, if you have to cross post make sure you have all of the
newsgroups in to TO: statement before you send it. It stops you from
wasting other peoples time not knowing someone else had resolved your
problem already in another group!

First off, you should keep the date in the database the default format
'yyyy-MM-DD' as a DATE Type..

I am not handling leap years but gives you a start. You may want to
build a procedure to do all of this and include leap year:

$minAge = 24;
$maxAge = 54;

SELECT
LASTNAME,
FIRSTNAME,
ROUND(DATEDIFF(CURDATE(),birthday)/365) AS 'numberOfYears',
birthday
FROM table
WHERE
ROUND(DATEDIFF(CURDATE(),birthday)/365) <= $maxAge AND
ROUND(DATEDIFF(CURDATE(),birthday)/365) >= $minAge;


--
Thanks in Advance...
IchBin, Pocono Lake, Pa, USA http://weconsultants.phpnet.us
__________________________________________________ ________________________

'If there is one, Knowledge is the "Fountain of Youth"'
-William E. Taylor, Regular Guy (1952-)