Print rows that don't exist in table?

I have a query that sums rows grouped by date. Basically, it counts how many membership applications in a day and prints out the results day by day. Naturally, if there are no applications in a day, it doesn't show that row (example below 15th and 17th had no applications.....

Date         |    Applications
-----------------------------------
12/01/2010   |   44
13/01/2010   |   73
14/01/2010   |   66
16/01/2010   |   102
18/01/2010   |   12
19/01/2010   |   22

I need it to print out the date and 0 for days when there are no applications, so as no days are skipped. Any idea how to do that. I thought of joining to a table with 开发者_C百科every day of the year, but that seems overkill

The query is below

SELECT
application_date AS Date,
COUNT(*) AS Applications 
FROM members
GROUP BY ap_date 

This is very similar to another question on SO. The general consensus seems to be:

1. Fill in the blank dates in the application layer of the program.
2. Create a temporary table, fill it with dates, and join on the temp table.

The best option is #1 - it's the least convoluted and should have the lowest overhead.

Select on the date range in question. Then iterate through the date range, looking to see if the current row matches. If it doesn't then output a 0. If it does then output the appropriate number and advance to the next row.

I would create a PHP array, and the array index would be a string, the date, I prefer this format YYYY-MM-DD and I would do something like this (note the key's date format is important)

// how far in the past to go
$days_in_the_past = 365;
// default to today, unix timestamp file
$start_day = time();

// the array that contains all out dates
$dates = array();

// sec * min * hours
$secs_in_a_day = 60 * 60 * 24;

// note we're going backwards here
for ( $i=0; $i <= $days_in_the_past; $i++ )
{
    $key = date('Y-M-D', ($start_day - ($secs_in_a_day * $i)));
    // this sets the format of   2010-01-21   as the key
    $dates[$key] = "";
}

$query = 'SELECT date, app FROM table';
$result = mysql_query($query);
while ( $row = mysql_fetch_assoc($result) )
{
    $key = date('Y-M-D', strtotime($row['Date']));
    $dates[] = $row['Applications'];
}

If you want the dates in order, just sort the array.

It's not overkill to create a table just for listing all the dates in the year; it's called a helper table, and can make this query (and lots like it) very easy.