我正在使用Google Chart API构建产品销售报告的折线图,并且每天使用UNION ALL作为粘合剂,基本上在每天的循环中运行相同的查询来分隔数据.
例如,一年的报告将使用UNION ALL 365次重复查询.
我猜这不是最好的做法 - 例如,如果有人会如此善良并指出我如何优化这个查询的正确方向,我将非常感激.
SELECT SUM(op.qty) AS qty
FROM uc_order_products op
LEFT JOIN uc_orders o ON o.order_id = op.order_id
LEFT JOIN node n ON n.nid = op.nid
LEFT JOIN node_type nt ON nt.type = n.type
WHERE (o.created > 1247695200) AND (o.created < 1247781600)
AND (o.order_status = 'completed')
AND (nt.type = 'book' OR nt.type = 'digital_movie')
UNION ALL
SELECT SUM(op.qty) AS qty
FROM uc_order_products op
LEFT JOIN uc_orders o ON o.order_id = op.order_id
LEFT JOIN node n ON n.nid = op.nid
LEFT JOIN node_type nt ON nt.type = n.type
WHERE (o.created > 1247781600) AND (o.created < 1247868000)
AND (o.order_status = 'completed')
AND (nt.type = 'book' OR nt.type = 'digital_movie')
UNION ALL
SELECT SUM(.......
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感谢所有快速回复!使用jspcal的查询示例,我的查询结果如下:
SELECT SUM(op.qty) AS qty, DATE_FORMAT(FROM_UNIXTIME(o.created),'%d-%m-%Y') AS day
FROM uc_order_products op
LEFT JOIN uc_orders o ON o.order_id = op.order_id
LEFT JOIN node n ON n.nid = op.nid
LEFT JOIN node_type nt ON nt.type = n.type
WHERE (o.created > 1247695200) AND (o.created < 1263596400)
AND (o.order_status = 'completed')
AND (nt.type = 'book' OR nt.type = 'digital_movie')
GROUP BY day
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使用PHP,我结合了一个完整的日期范围数组(匹配数组键与strtotime($data->day)查询结果),以获得没有销售的日子.
您可以按天分组:
select sum(op.qty) as qty, date_format(o.created, '%Y-%m-%d') as day
from ... where ... o.created >= subdate(now(), interval 1 year) and
o.created <= now() group by day order by day
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将报告数据集中一年中每一天的总和
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