Question

我有一个查询，显示该月最后一天的所有匹配报告（即完整月份报告）

SELECT  id, 
    DATE_FORMAT( start_date,  '%e-%b-%y' ) AS start_date, 
    DATE_FORMAT( end_date,  '%e-%b-%y' ) AS end_date, 
    DATE_FORMAT( end_date,  '%M %Y' ) as report_month
FROM Reports
WHERE customer_id ='2321'
  AND end_of_month_file = '1'

问题是重复的 - 有些文件已被多次处理......

id start_date  end_date   report_month
1  1-Feb-14    28-Feb-14  February 2014
2  1-Mar-14    31-Mar-14  March 2014
3  1-Mar-14    31-Mar-14  March 2014
4  1-Mar-14    31-Mar-14  March 2014
5  1-Mar-14    31-Mar-14  March 2014
6  1-Mar-14    31-Mar-14  March 2014
7  1-Mar-14    31-Mar-14  March 2014
8  1-Mar-14    31-Mar-14  March 2014
9  1-Mar-14    31-Mar-14  March 2014

如何过滤结果，只给我每个月的最后一个/最新条目？

Answer 1

使用GROUP BY然后排序最新的：

SELECT  id, 
    DATE_FORMAT( start_date,  '%e-%b-%y' ) AS start_date, 
    DATE_FORMAT( end_date,  '%e-%b-%y' ) AS end_date, 
    DATE_FORMAT( end_date,  '%M %Y' ) as report_month
FROM Reports
WHERE customer_id ='2321'
  AND end_of_month_file = '1'
GROUP BY start_date
ORDER BY start_date DESC;

Answer 2

既然你有MySQL，为什么不使用LIMIT：

SELECT  id, 
    DATE_FORMAT( start_date,  '%e-%b-%y' ) AS start_date, 
    DATE_FORMAT( end_date,  '%e-%b-%y' ) AS end_date, 
    DATE_FORMAT( end_date,  '%M %Y' ) as report_month
FROM Reports
WHERE customer_id ='2321'
  AND end_of_month_file = '1'
ORDER BY start_date DESC
LIMIT 1

选择所有匹配的行但仅显示最新的

2 个答案: