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,$mon th,$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!]

Re: I need a genius for this one!

In message <1158561685.666 825.252900@h48g 2000cwc.googleg roups.com>,
ameshkin <amir.meshkin@g mail.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,$mon th,$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/en/date-and-time-functions.html

Check the YEAR function.



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

[Mitul]

Re: I need a genius for this one!

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($year,$mon th,$day) = explode("-",$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_D ATE)-YEAR(birth))
-- (RIGHT(CURRENT_ DATE,5)<RIGHT(b irth,5))
-AS age
-FROM pet;


/***** There is nothing impossible coz Impossible itself means I M
Possible. :)
Regards,
Mitul Patel

[Oli Filth]

Re: I need a genius for this one!

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($year,$mon th,$day) = explode("-",$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/en/date-and-time-functions.html

--
Oli

[ameshkin]

Re: I need a genius for this one!

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,$mon th,$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/en/date-and-time-functions.html

>
>
--
Oli

[IchBin]

Re: I need a genius for this one!

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($year,$mon th,$day) = explode("-",$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(),birth day)/365) AS 'numberOfYears' ,
birthday
FROM table
WHERE
ROUND(DATEDIFF( CURDATE(),birth day)/365) <= $maxAge AND
ROUND(DATEDIFF( CURDATE(),birth day)/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]

Re: I need a genius for this one!

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($year,$mon th,$day) = explode("-",$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(),birth day)/365) AS 'numberOfYears' ,
birthday
FROM table
WHERE
ROUND(DATEDIFF( CURDATE(),birth day)/365) <= $maxAge AND
ROUND(DATEDIFF( CURDATE(),birth day)/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-)