Presto 滞后日期，group/partitioned by id

Question

假设我想在每次客户更新预算时查找。

我的数据是这样的

datetime, client_id, new_budget
__________,__________,___________
2022-01-01,1,        ,100
2022-01-01,2,        ,300
2022-01-02,1,        ,80
2022-01-02,2,        ,80

还有我的代码运行。

SELECT datetime AS dt_1,
       LAG(datetime) OVER (ORDER BY client_id, datetime) AS dt_2,
       client_id,
       new_budget
FROM budget_table

我期望返回的是

dt_1,      dt_2,      client_id, new_budget
__________,__________,__________,___________
2022-01-01,NULL,      1         , 100
2022-01-02,2022-01-01,1         , 80
2022-01-01,NULL,      2         , 300
2022-01-02,2022-01-01,2         , 80

因此在每个 client_id 的第一个条目中 dt_2 有 NULL 值。我不确定什么代码可以实现这种效果；是否需要 GROUP BY 子句（或 partition over 子句）。

但这是我运行

SQL 的输出

dt_1,      dt_2,      client_id, new_budget
__________,__________,__________,___________
2022-01-01,NULL,      1         , 100
2022-01-02,2022-01-01,1         , 80
2022-01-01,2022-01-02,2         , 300
2022-01-02,2022-01-01,2         , 80

所以这里的大问题是，如果前一行来自不同的 client_id.

，它无法识别 dt_2 应该为 NULL

建议使用哪种语法来实现此效果？

Answer 1

你需要partition by client_id:

The PARTITION BY clause separates the input rows into different partitions. This is analogous to how the GROUP BY clause separates rows into different groups for aggregate functions. If PARTITION BY is not specified, the entire input is treated as a single partition.

SELECT datetime AS dt_1,
       LAG(datetime) OVER (PARTITION BY client_id ORDER BY datetime) AS dt_2,
       client_id,
       new_budget
FROM budget_table

Presto 滞后日期，group/partitioned by id

Presto lag dates, group/partitioned by id

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