Question

我有一个如下所示的数据集：

    respondent_id   day_session daydiff
nmo876  11/19/2017  0
nmo876  11/20/2017  1
nmo876  11/21/2017  1
nmo876  11/23/2017  2
nmo876  11/24/2017  1
nmo876  11/25/2017  1
nmo876  11/26/2017  1
nmo876  11/27/2017  1
nmo876  11/28/2017  1
nmo876  11/29/2017  1
nmo876  11/30/2017  1
nmo876  12/1/2017   1
nmo876  12/2/2017   1
nmo876  12/3/2017   1
nmo876  12/4/2017   1
nmo876  12/5/2017   1
nmo876  12/6/2017   1
nmo876  12/7/2017   1
nmo876  12/8/2017   1
nmo876  12/9/2017   1
nmo876  12/10/2017  1
nmo876  12/11/2017  1
nmo876  12/12/2017  1
nmo876  12/13/2017  1
nmo876  12/14/2017  1
nmo876  12/15/2017  1
nmo876  12/16/2017  1
nmo876  12/17/2017  1
nmo876  12/18/2017  1
nmo876  12/19/2017  1
nmo876  12/20/2017  1
nmo876  12/23/2017  3
nmo876  12/24/2017  1
nmo876  12/26/2017  2
nmo876  12/27/2017  1
nmo876  12/28/2017  1
nmo876  12/29/2017  1
nmo876  12/30/2017  1
nmo876  12/31/2017  1
nmo876  1/2/2018    2
nmo876  1/3/2018    1
nmo876  1/4/2018    1
nmo876  1/5/2018    1

我想编写一个脚本，从用户可能有多个连续day_sessions的数据集中选择最大的连续day_sessions块，即daydiff = 1.对于nmo876，输出为27。

以下是代码应该计算连续每日会话的最大块的更多数据：

jkl567  11/19/2017  1
jkl567  11/20/2017  1
jkl567  11/21/2017  1
jkl567  11/22/2017  1
jkl567  11/23/2017  1
jkl567  11/24/2017  1
jkl567  11/25/2017  1
jkl567  11/26/2017  1
jkl567  11/27/2017  1
jkl567  11/28/2017  1
jkl567  11/29/2017  1
jkl567  11/30/2017  1
jkl567  12/1/2017   1
jkl567  12/2/2017   1
jkl567  12/3/2017   1
jkl567  12/4/2017   1
jkl567  12/5/2017   1
jkl567  12/6/2017   1
jkl567  12/7/2017   1
jkl567  12/8/2017   1
jkl567  12/9/2017   1
jkl567  12/10/2017  1
jkl567  12/11/2017  1
jkl567  12/12/2017  1
jkl567  12/13/2017  1
jkl567  12/14/2017  1
jkl567  12/15/2017  1
jkl567  12/16/2017  1
jkl567  12/17/2017  1
jkl567  12/18/2017  1
jkl567  12/19/2017  1
jkl567  12/20/2017  1
jkl567  12/21/2017  1
jkl567  12/22/2017  1
jkl567  12/23/2017  1
jkl567  12/24/2017  1
jkl567  12/25/2017  1
jkl567  12/26/2017  2
jkl567  12/28/2017  1
jkl567  12/29/2017  3
jkl567  1/1/2018    1
jkl567  1/2/2018    1
jkl567  1/3/2018    1
jkl567  1/4/2018    1

Answer 1

您可以使用变量来实现此目的。

SQL> with test (fullname) as
  2    (select 'Metro Goldwyn Mayer' from dual
  3    )
  4  select substr(fullname, 1, instr(fullname, ' ', 1, 1) - 1) first_name,
  5         substr(fullname,
  6                instr(fullname, ' ', 1, 1) + 1,
  7                instr(fullname, ' ', 1, 2) - instr(fullname, ' ', 1, 1) - 1) middle_name,
  8         substr(fullname,
  9                instr(fullname, ' ', 1, 2) + 1) last_name
 10  from test;

FIRST_NAME           MIDDLE_NAME          LAST_NAME
-------------------- -------------------- --------------------
Metro                Goldwyn              Mayer

SQL>

<强> 解释

内部查询使用附加列作为计数器准备结果，仅当daydiff为1时才会递增。

外部查询然后合并结果并显示最高值。

Fiddle

计算MySQL文件中最大的连续会话块

1 个答案: