display month then day for a given year

[[deleted]]

What I'm trying to accomplish is the following:

2003

July
5- Result
7- Result

August
10- Result
12- Result

Currently what I have is the following:

$space = ' '; 
	$cur_year= date('Y');
	$result = mysql_query("SELECT DISTINCT i.racedate,r.racename FROM 
							crc1.race r, crc1.raceindex i
 							WHERE r.raceid = i.race_id AND date_format(i.racedate,'%Y') = '$cur_year'");

		if (mysql_num_rows($result)>0){
 			echo '<b>'.date('Y').'</b><br/><br/>';
		while ($row= mysql_fetch_array($result)){
 			echo '<b>'.date("F",strtotime($row['racedate'])).'</b>';
 			echo '<br/>';
 			echo date("d",strtotime($row['racedate']));echo $space;echo $row['racename'];
 			echo '<br/><br/>';
		 }//end while
	}//end if

This displays the results as:

August
10- Result

August
11- Result

How do I correct this?

[Pig]

You have to do 2 queries

1. get all months
2. print month
3. for each month, get all days
4. print all days

   $result_month = mysql_query("something here to get the month");
   
   while( $row_month = mysql_fetch_array($result_month) )
   {
        print("$row_month[month]");
        $result_day = mysql_query("something here to get the days");
        while( $row_day = mysql_fetch_array($result_day) )
        {
             echo("$row_day[day]<br>");
        }
        echo("<br>");
   }
   

[[deleted]]

Ok,but I don't want all months or all days. I only want months where there are days with data.

[tyr_82]

Check out the mysql-documentation about pattern-matching at http://www.mysql.com/documentation/mysql/bychapter/manual_Tutorial.html#Pattern_matching
and read more about it and how to use.