show result from percentage - php

Question

Hello I have I field in my table called percentage but what I looking to do is only show a result based on percentage ie if I set 10 it would only show the result 10% of the time.
here is my database which is Joomla database code in mysql
// Select the required fields from the table.
$query->select(
'a.id, a.title, a.simulator, a.generation, a.quality, a.quantity, a.percentage, a.attribute, a.checked_out, a.checked_out_time, a.catid' .
', a.state, a.access, a.created, a.created_by, a.version, a.ordering'
);
$query->from('#__evolutionary_texture AS a');
// Join over the users for the checked out user.
$query->select('uc.name AS editor')
->join('LEFT', '#__users AS uc ON uc.id=a.checked_out');
// Join over the asset groups.
$query->select('ag.title AS access_level')
->join('LEFT', '#__viewlevels AS ag ON ag.id = a.access');
// Join over the species.
$query->select('category.title AS category_title')
->join('LEFT', '#__categories AS category ON category.id = a.catid');
$query->select('parent.title AS parent_title')
->join('left', '#__categories AS parent ON parent.id=category.parent_id');
// Join over the section.
//$query->select('section.title AS section_title')
// ->join('LEFT', '#__categories AS section ON section.id = a.catid');
// Join over the users for the author.
$query->select('ua.name AS author_name')
->join('LEFT', '#__users AS ua ON ua.id = a.created_by');
$query->where('parent.title = ' . $db->quote($species));
//echo "section = $section";
$query->where('category.title = ' . $db->quote($section));
// check where generation.
$query->where('a.generation = ' . $db->quote($generation));
// check where simulator.
$query->where('a.simulator = ' . $db->quote($simulator));
// Debug the query
//var_dump($db->replacePrefix( (string) $query ) );
// Set the query and load the result.
$db->setQuery($query);
as what I trying to do is only show texture based on the percentage as this data is imported into a game called second life
my percentage field right now goes from 0 - 100

Answer 1

After analyzing your code/problem, I can suggest following.
you need to store a variable random_no and times somewhere, in SESSION, DATABSE or APPLIATION level var
The value of random_no will be =array() of 100 int values from 1 to 100 randomly// example =array(23,15,67,98, ...., 5); // all 1-100 numbers once only
value of times will be 0 initially and will increase by 1 after each query. When value reach to 99 then next it will be 0 and recalculate the value of random_no. //code at where are querying database
Now run your query with condition: a.percentage >= $random_no[$times]
creating $randome_no array:
$random_no=array();
for($i=1;$i<=100;i++){
$random_no[]=$i;
}
shuffle($random_no);
PHP shuffle
Above method will select records percentage times randomly