Jim*_*y M 4 php mysql date sql-order-by
我在我的数据库中使用以下查询,
SELECT b.sales_id,b.category_id,b.sale_starts,b.sale_ends
FROM tbl_sales b WHERE b.active=1
UNION
SELECT b.sales_id,b.category_id,b.sale_starts,b.sale_ends
FROM tbl_sales b INNER JOIN tb_category c ON b.category_id=c.cat_id
WHERE c.cat_keyword LIKE 'a' ORDER BY sale_ends DESC
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得到如下结果,
sales_id | category_id |sale_starts | sale_ends
----------|---------------------|------------|--------------
1 | 10 | 2012-03-31 | 2012-04-30
2 | 11 | 2012-03-22 | 2012-04-27
3 | 25 | 2012-03-31 | 2012-04-25
4 | 12 | 2012-04-05 | 2012-04-11
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现在我需要得到如下结果,即具有today's date/current date as
sale_ends 的行必须显示在订单的顶部(假设今天的日期/当前日期是2012-04-11
),如下所示 -
sales_id | category_id |sale_starts | sale_ends
----------|---------------------|------------|--------------
4 | 12 | 2012-04-05 | 2012-04-11 (today's date)
1 | 10 | 2012-03-31 | 2012-04-30
2 | 11 | 2012-03-22 | 2012-04-27
3 | 25 | 2012-03-31 | 2012-04-25
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需要帮助,谢谢你提前.
尝试使用条件的ORDER BY子句 -
ORDER BY IF(sale_ends = DATE(NOW()), 0, 1), sale_ends DESC
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