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这是我的原始查询.... 如果您可以看到 next_row_date '2013-01-01' 的日期不是 fld_date 的最后一条记录，并且 next_row_date 的最后一条记录是 '2013-01-15' 必须为 0 .

+-----+---+------------+---------------+
| 序列 | 我 | fld_日期 | 下一行日期 |
+-----+---+------------+---------------+
| 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。你能帮我解决这个问题吗？

+-----+---+------------+---------------+
| 序列 | 我 | fld_日期 | 下一行日期 |
+-----+---+------------+---------------+
| 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;
4

1 回答 1

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：

| 序列 | 我 | FLD_DATE | 下一个日期 |
-------------------------------------
| 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：即使您在示例数据 47-49 中的 seq 编号中存在间隙，查询也将起作用

于 2013-06-27T01:46:29.070 回答