在 ID 列 + select prev/next 日期列中找到差距

Question

我们有一个 table 和 ID 带间隙的自动增量列。每行还有一个 created 日期。

我们正在尝试为每个间隔 ID（缺失的）找出我们 table 中可用的 prev/next 日期。

我们已经构建了一个 SQL 来识别间隙（来自 here 的解决方案），所以我将它们缓存在 table 中，但是从这里如何找到 prev/next created 日期弥补了原来 table 的差距。

输入：

+----+------------+
| 84 | 1443728132 |
| 91 | 1443728489 |
| 93 | 1443729058 |
| 94 | 1443729200 |
+----+------------+

输出

+--------+------------+------------+
| gap_id |  prev_dt   |  next_dt   |
+--------+------------+------------+
|     85 | 1443728132 | 1443728489 |
|     86 | 1443728132 | 1443728489 |
|     87 | 1443728132 | 1443728489 |
|     88 | 1443728132 | 1443728489 |
|     89 | 1443728132 | 1443728489 |
|     90 | 1443728132 | 1443728489 |
|     92 | 1443728489 | 1443729058 |
+--------+------------+------------+

我最后做了 Google BigQuery。

Answer 1

应该适用于 MySQL

以外的大多数数据库

select      *

from       (select      lag (id) over (order by id) + 1                                 as gap_start
                       ,id - 1                                                          as gap_end
                       ,lag (dt) over (order by id)                                     as dt_before_gap
                       ,dt                                                              as dt_after_gap
                       ,case when lag (id) over (order by id) + 1 <> id then 'Y' end    as is_gap

            from        t
            ) t

where       is_gap = 'Y'
;

Answer 2

我建议将结果分组到范围内：

select min(id + 1) as first_missing_id,
       (next_id - 1) as last_missing_id,
       next_dte
from (select t.*,
             lead(id) over (order by id) as next_id,
             lead(dte) over (order by id) as next_dte
      from t
     ) t
where next_id <> id + 1
group by next_id, next_dte;

获取个人 ID 很棘手。毕竟，如果您有 1, 1000000, 1000000000 那么您可能会生成很多行。

Answer 3

对于 BigQuery 标准 SQL

WITH yourTable AS (
SELECT 84 AS id, 1443728132 AS dt UNION ALL
SELECT 91 AS id, 1443728489 AS dt UNION ALL
SELECT 93 AS id, 1443729058 AS dt UNION ALL
SELECT 94 AS id, 1443729200 AS dt 
),
nums AS (
  SELECT num 
  FROM UNNEST(GENERATE_ARRAY((SELECT MIN(id) FROM YourTable), (SELECT MAX(id) FROM YourTable))) AS num
),
gaps AS (
  SELECT
    LAG (id) OVER (ORDER BY id) + 1 AS gap_start,
    id - 1 AS gap_end,
    LAG (dt) OVER (ORDER BY id) AS prev_dt,
    dt AS next_dt,
    CASE 
      WHEN LAG (id) OVER (ORDER BY id) + 1 <> id THEN 'Y'
    END AS is_gap
  FROM
    yourTable 
)
SELECT num as gap_id, prev_dt, next_dt
FROM gaps JOIN nums 
ON num BETWEEN gap_start AND gap_end
WHERE is_gap = 'Y'
ORDER BY num

输出：

gap_id     prev_dt     next_dt   
85      1443728132  1443728489   
86      1443728132  1443728489   
87      1443728132  1443728489   
88      1443728132  1443728489   
89      1443728132  1443728489   
90      1443728132  1443728489   
92      1443728489  1443729058

在 ID 列 + select prev/next 日期列中找到差距

Find the gap in ID column + select the prev/next date column

sql

google-bigquery

gaps-and-islands