Практика: общие табличные выражения. Задача 2

WITH i_month AS
  (SELECT EXTRACT(MONTH FROM CAST(invoice_date AS DATE)) AS invoice_month
   FROM invoice
   GROUP BY invoice_month
   ORDER BY invoice_month),
inv_2011 AS
  (SELECT EXTRACT(MONTH FROM CAST(invoice_date AS DATE)) AS invoice_month,
          COUNT(invoice_id) AS year_2011
   FROM invoice
   WHERE EXTRACT(YEAR FROM CAST(invoice_date AS DATE)) = 2011
   GROUP BY invoice_month),
inv_2012 AS
  (SELECT EXTRACT(MONTH FROM CAST(invoice_date AS DATE)) AS invoice_month,
          COUNT(invoice_id) AS year_2012
   FROM invoice
   WHERE EXTRACT(YEAR FROM CAST(invoice_date AS DATE)) = 2012
   GROUP BY invoice_month),
inv_2013 AS
  (SELECT EXTRACT(MONTH FROM CAST(invoice_date AS DATE)) AS invoice_month,
          COUNT(invoice_id) AS year_2013
   FROM invoice
   WHERE EXTRACT(YEAR FROM CAST(invoice_date AS DATE)) = 2013
   GROUP BY invoice_month)


SELECT i_month.invoice_month,
       inv_2011.year_2011,
       inv_2012.year_2012,
       inv_2013.year_2013
FROM
  (SELECT EXTRACT(MONTH FROM CAST(invoice_date AS DATE)) AS invoice_month
   FROM invoice
   GROUP BY invoice_month
   ORDER BY invoice_month) AS i_month

LEFT JOIN
  inv_2011 ON i_month.invoice_month = inv_2011.invoice_month

LEFT JOIN
  inv_2012 ON i_month.invoice_month = inv_2012.invoice_month

LEFT JOIN
  inv_2013 ON i_month.invoice_month = inv_2013.invoice_month;