这是我的原始查询.... 如果您可以看到next_row_date'2013-01-01'的日期,它不是fld_date的最后一条记录,而next_row_date的最后一条记录是'2013-01-15',则必须为0。
+-----+---+------------+---------------+ | seq | i | fld_Date | next_row_date | +-----+---+------------+---------------+ | 41 | 1 | 2012-10-08 | 2012-10-15 | | 42 | 2 | 2012-10-15 | 2012-10-22 | | 43 | 3 | 2012-10-22 | 2012-10-29 | | 44 | 4 | 2012-10-29 | 2012-11-05 | | 45 | 5 | 2012-11-05 | 2012-11-12 | | 46 | 6 | 2012-11-12 | 2012-11-19 | | 47 | 7 | 2012-11-19 | 2013-01-01 | | 49 | 8 | 2013-01-08 | 2013-01-15 | +-----+---+------------+---------------+
这是我想要的输出。你可以用这个查询来帮助我吗?
+-----+---+------------+---------------+ | seq | i | fld_Date | next_row_date | +-----+---+------------+---------------+ | 41 | 1 | 2012-10-08 | 2012-10-15 | | 42 | 2 | 2012-10-15 | 2012-10-22 | | 43 | 3 | 2012-10-22 | 2012-10-29 | | 44 | 4 | 2012-10-29 | 2012-11-05 | | 45 | 5 | 2012-11-05 | 2012-11-12 | | 46 | 6 | 2012-11-12 | 2012-11-19 | | 47 | 7 | 2012-11-19 | 2013-01-08 | | 49 | 8 | 2013-01-08 | 0 | +-----+---+------------+---------------+
SELECT
db_lms.a.seq,
(@i:=@i+1)AS i,
db_lms.a.fld_Date,
(db_lms.b.fld_Date)AS next_row_date
FROM db_lms.lms_savings a, db_lms.lms_savings b, (SELECT @i:=0) ii
WHERE (db_lms.a.seq = db_lms.b.seq-1) ORDER BY db_lms.a.seq ASC;
答案 0 :(得分:1)
尝试
SELECT seq, @i := @i + 1 i, fld_Date, next_date
FROM
(
SELECT seq, @d next_date, @d := fld_Date fld_Date
FROM lms_savings, (SELECT @d := 0) d
ORDER BY seq DESC
) q, (SELECT @i := 0) n
ORDER BY seq
输出:
| SEQ | I | FLD_DATE | NEXT_DATE | ------------------------------------- | 41 | 1 | 2012-10-08 | 2012-10-15 | | 42 | 2 | 2012-10-15 | 2012-10-22 | | 43 | 3 | 2012-10-22 | 2012-10-29 | | 44 | 4 | 2012-10-29 | 2012-11-05 | | 45 | 5 | 2012-11-05 | 2012-11-12 | | 46 | 6 | 2012-11-12 | 2012-11-19 | | 47 | 7 | 2012-11-19 | 2013-01-08 | | 49 | 8 | 2013-01-08 | 0 |
这是 SQLFiddle 演示。
注意:即使您的样本数据中存在空白,查询仍然有效,例如样本数据47-49