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Mysql zur Auswahl der monatlichen Aufzeichnung, auch wenn keine Daten vorhanden sind

Ich werde nicht viel über die Effizienz sagen, da ich es nicht mit anderen Methoden getestet habe, aber ohne eine temporäre Tabelle zu haben, scheint dies ein guter Weg zu sein.

   SELECT COUNT(u.userID) AS total, m.month
     FROM (
           SELECT 'Jan' AS MONTH
           UNION SELECT 'Feb' AS MONTH
           UNION SELECT 'Mar' AS MONTH
           UNION SELECT 'Apr' AS MONTH
           UNION SELECT 'May' AS MONTH
           UNION SELECT 'Jun' AS MONTH
           UNION SELECT 'Jul' AS MONTH
           UNION SELECT 'Aug' AS MONTH
           UNION SELECT 'Sep' AS MONTH
           UNION SELECT 'Oct' AS MONTH
           UNION SELECT 'Nov' AS MONTH
           UNION SELECT 'Dec' AS MONTH
          ) AS m
LEFT JOIN users u 
ON MONTH(STR_TO_DATE(CONCAT(m.month, ' 2013'),'%M %Y')) = MONTH(u.userRegistredDate)
   AND YEAR(u.userRegistredDate) = '2013'
GROUP BY m.month
ORDER BY 1+1;

Wenn Sie die Vereinigung basierend auf einem Datumsformat erstellen, können Sie sogar die Arbeit und Belastung der Abfrage reduzieren.

   SELECT COUNT(u.userID) AS total, DATE_FORMAT(merge_date,'%b') AS month, YEAR(m.merge_date) AS year
     FROM (
           SELECT '2013-01-01' AS merge_date
           UNION SELECT '2013-02-01' AS merge_date
           UNION SELECT '2013-03-01' AS merge_date
           UNION SELECT '2013-04-01' AS merge_date
           UNION SELECT '2013-05-01' AS merge_date
           UNION SELECT '2013-06-01' AS merge_date
           UNION SELECT '2013-07-01' AS merge_date
           UNION SELECT '2013-08-01' AS merge_date
           UNION SELECT '2013-09-01' AS merge_date
           UNION SELECT '2013-10-01' AS merge_date
           UNION SELECT '2013-11-01' AS merge_date
           UNION SELECT '2013-12-01' AS merge_date
          ) AS m
LEFT JOIN users u 
       ON MONTH(m.merge_date) = MONTH(u.userRegistredDate)
          AND YEAR(m.merge_date) = YEAR(u.userRegistredDate)
GROUP BY m.merge_date
ORDER BY 1+1;

Live-DEMO beider Abfragen.