Sorting by age issue.

Jako · 07-31-2008, 05:08 PM

I have a table that stores users birthdays, date, month year, etc.

I then have a function which turns these into an age, I.E. 24.

Problem is when I want to show all users who are a specific age I get stuck. I can do it an easy way and sort by their year born, but this will sometimes show a fluctuation because everyone born from 1988 may not be 20 yet.

Is there any work around for this?

Karl · 07-31-2008, 05:50 PM

You could use MySQL to work out their age. However, the MySQL for that is rather intensive, and would require the HAVING clause. How about simply returning them all by the year, and then in the loop simply calculate their age and if it doesn't match ignore that one and continue.

Jako · 07-31-2008, 07:20 PM

Quote:

Originally Posted by Karl

You could use MySQL to work out their age. However, the MySQL for that is rather intensive, and would require the HAVING clause. How about simply returning them all by the year, and then in the loop simply calculate their age and if it doesn't match ignore that one and continue.

Thanks for the reply, I'm just learning PHP along the way and I'm not the best at while loops.

How would I go about setting it up?

The number I am going to pass through will be the age then, so

$id=$_GET['id']; would be 23 for example.

And the variables are $month, $day and $year which when put in calculate_age($month, $day, $year) will calculate the age.

Not too sure how to write the while loop to look for the specified age.

Ross · 07-31-2008, 07:27 PM

You'll use this a lot when dealing with multiple rows of data:

PHP Code:


			
$query = mysql_query('SELECT * FROM `people`');

// Check at leats one row has been returned
if(mysql_num_rows($query) < 1)
{
    exit;
}

/**
 * $query is a resource that contains multiple rows of data, you can 
 * use fetch_assoc or fetch_array to get this data
 */
 
/**
 * While loops keep going while the condition is true, or in this case while there are still rows in $query
 */
$ages = array();

while($row = mysql_fetch_assoc($query))
{
    // Add a new element to the array with the age
    $ages[] = calculate_age($row['day'], $row['month'], $row['year']);
}

Jako · 07-31-2008, 07:42 PM

PHP Code:


			
$ages = array();



while($row = mysql_fetch_assoc($result))

{

    // Add a new element to the array with the age

    $ages[] =  calculate_age($month, $day, $year);

}

Ok so I condensed it down to what I think I needed, but how do I input the age into there.

Like I said above, if $id = 23, how do I only display the results from the calculation function which are 23.

Jako · 07-31-2008, 08:00 PM

I ended up getting it, finally figured it out haha but thanks to all for your help.

After I built the array, I used the in_array() function to find the $id which is the age I was looking for to output my results.

thanks.

Ross · 07-31-2008, 09:34 PM

Quote:

Originally Posted by Jako

I ended up getting it, finally figured it out haha but thanks to all for your help.

After I built the array, I used the in_array() function to find the $id which is the age I was looking for to output my results.

thanks.

Were you only looking to get the age of person.id #23? IF so you're much better off doing something like this:

PHP Code:


			
$query = mysql_query('SELECT * FROM `people` WHERE `id` = 23 LIMIT 1');



$person = mysql_fetch_assoc($query);

// Get the data from this array as in the while loop - but you don't need one for this example



print_r($person);

It's much more efficient and easier if you only need one person's results.

Jako · 08-02-2008, 05:47 AM

This is what I ended up going with which worked perfectly

PHP Code:


			
<?php 
while($row = mysql_fetch_assoc($result)) {
$ages = array();
$ages[] = calculate_age($row['month'], $row['day'], $row['year']);
if (in_array($a, $ages)) { ?>

and $a was the age I was passing through the url.