复杂SQL查询实例-5种普惠产品必须显示...

复杂SQL需求：

1.查询productCode in (1, 2, 4, 5, 7)五种

2.5种产品必须固定显示，优先显示procuct_status='1'在售产品，在售产品卖完则售罄产品顶上来，即每种产品必须显示一条，不可空挡

3.在售取时间最早的一条，售罄取时间最晚的一条

SELECT
	procuct_starttime,
	procuct_starttime AS "product_saletime",
	product_no,
	o.product_code,
	product_name,
	marketing_type,
	pay_type,
	is_transfer,
	close_period,
	procuct_status,
	product_recommend,
	trade_type,
	IFNULL(
		(
			SELECT
				t.star_img
			FROM
				tb_crm_activity_product_info f,
				tb_crm_activity_resource_info g,
				tb_crm_star_information t
			WHERE
				product_no = f.product_id
			AND f.acti_id = g.acti_id
			AND g.resource_id = t.id
		),
		'http://www.gougou.com.cn'
	) url,
	IFNULL(
		(
			SELECT
				COUNT(*)
			FROM
				tb_customer_order co
			WHERE
				co.product_no = o.product_no
			GROUP BY
				co.product_no
		),
		0
	) buyCount,
	IFNULL(procuct_aroe, 0) annualizedRate,
	product_addonrate,
	IFNULL(procuct_min_amount, 0) minPurchaseAmount,
	procuct_amount financingAmount,
	IFNULL(product_amount_buy, 0) currentAmount,
	FORMAT(
		IFNULL(product_amount_buy, 0) * 100 / procuct_amount,
		2
	) amountScale,
	CASE
WHEN FORMAT(
	IFNULL(product_amount_buy, 0) * 100 / procuct_amount,
	2
) >= 100 THEN
	'1'
ELSE
	'0'
END eee,
 procuct_type productType,
 procuct_day productDay,
 procuct_summary productSummary,
 CASE
WHEN product_amount_buy >= procuct_amount THEN
	1
ELSE
	0
END oindex,

IF (
	o.product_recommend = 'Y',
	1,
	0
) ooo
FROM
	(
		SELECT
			pt.*
		FROM
			tb_product_type pt
		INNER JOIN (
			SELECT
				MAX(product_no) product_no
			FROM
				tb_product_type a
			INNER JOIN (
				SELECT
					product_code,
					MIN(procuct_starttime) procuct_starttime
				FROM
					tb_product_type
				WHERE
					marketing_type = '1'
				AND procuct_status = '1'
				AND product_amount_buy < procuct_amount
				AND product_code IN (1, 2, 4, 5, 7)
				GROUP BY
					product_code
			) b ON a.product_code = b.product_code
			AND a.procuct_starttime = b.procuct_starttime
			GROUP BY
				a.product_code
			UNION ALL
				SELECT
					MAX(product_no) product_no
				FROM
					tb_product_type a
				INNER JOIN (
					SELECT
						p1.product_code,
						MAX(procuct_starttime) procuct_starttime
					FROM
						tb_product_type p1
					WHERE
						NOT EXISTS (
							SELECT
								1
							FROM
								tb_product_type p2
							WHERE
								marketing_type = '1'
							AND procuct_status = '1'
							AND product_amount_buy < procuct_amount
							AND p1.product_code = p2.product_code
						)
					AND marketing_type = '1'
					AND product_code IN (1, 2, 4, 5, 7)
					GROUP BY
						product_code
				) b ON a.product_code = b.product_code
				AND a.procuct_starttime = b.procuct_starttime
				GROUP BY
					a.product_code
		) pt2 ON pt.product_no = pt2.product_no
	) o
ORDER BY
	ooo DESC,
	product_code ASC;