使用 bigquery 查找特定日期的整体数据答案

【问题标题】：Finding the overall data from specific date using bigquery使用 bigquery 查找特定日期的整体数据
【发布时间】：2020-11-02 06:29:07
【问题描述】：

我正在使用 bigquery 创建一个查询，该查询将计算每个 ID 的特定日期起 5 天内的数据总数。到目前为止，我设法获得了结果，其中结果返回了每天（第 1 天到第 5 天）的数据，这意味着结果中有 5 个日期。但是，我想要的实际上是只返回 1 个日期，这是最后一个日期，所以它就像 1 个日期，显示每个 ID 从该特定日期起过去 5 天的整体数据。

样本数据：

示例代码：

  SELECT
   ID,
   Date,
    SUM(CASE WHEN Language = 'EN' THEN 1 ELSE 0 END) AS lang_EN,
    SUM(CASE WHEN Language = 'SN' THEN 1 ELSE 0 END) AS lang_SN,
     FROM (
      SELECT 
      DATE(Timestamp) as Date ,
       ID, 
      CASE 
     WHEN Language in ('EN', 'English') THEN 'EN' 
     WHEN Language in ('MY', 'Malay')  THEN 'MY' ELSE Language 
     END AS Language,  
           FROM t
           WHERE Smooch_User_ID IS NOT NULL AND DATE(Timestamp) between '2020-01-01' and '2020-01-31'
           GROUP BY   ID, Language, DATE(Timestamp) 
         )
          GROUP BY  ID,Date

样本输出：

到目前为止，查询返回从第 1 天到第 5 天的每个日期的数据。如何使它返回从第 1 天到第 5 天的整体数据，并仅返回第 5 天的日期列日期，如下所示（假设第 1 天是 2020-01-01，第 5 天是 2020-01-05）。

【问题讨论】：

你能分享你想要达到的期望输出的样本吗？
您应该展示您的初始输入数据以及所需的输出。你现在的问题在我看来是一个中间数据，无助于理解你的情况。见How to create a Minimal, Reproducible Example

标签： sql google-bigquery timestamp

【解决方案1】：

如果您使用脚本，我认为一种方法可能是声明变量，以便您可以在间隔中根据 end_date 提取/标记输出。例如。

    declare end_dt date default "2020-07-05"; -- this can be used as placeholder and replaced at run-time based on execution
    declare start_dt date default DATE_SUB(end_dt, INTERVAL 5 DAY);
    
    select 
      end_dt as ts_date, 
      id, 
      sum(CASE WHEN lang = 'EN' THEN 1 ELSE 0 END) AS lang_EN, 
      sum(CASE WHEN lang = 'SN' THEN 1 ELSE 0 END) AS lang_SN 
    from(
      -- sample data
      select cast("2020-07-01 01:01:25.550 UTC" as timestamp) as ts, "A" as id, "EN" as lang
      union all
      select cast("2020-07-01 01:01:25.550 UTC" as timestamp) as ts, "A" as id, "EN" as lang
      union all
      select cast("2020-07-01 01:01:25.550 UTC" as timestamp) as ts, "A" as id, "SN" as lang
      union all
      select cast("2020-07-02 01:01:25.550 UTC" as timestamp) as ts, "B" as id, "EN" as lang
      union all
      select cast("2020-07-02 01:01:25.550 UTC" as timestamp) as ts, "B" as id, "EN" as lang
      union all
      select cast("2020-07-02 01:01:25.550 UTC" as timestamp) as ts, "B" as id, "SN" as lang
      union all
      select cast("2020-07-02 01:01:25.550 UTC" as timestamp) as ts, "B" as id, "SN" as lang
      union all
      select cast("2020-07-03 01:01:25.550 UTC" as timestamp) as ts, "C" as id, "SN" as lang
      union all
      select cast("2020-07-03 01:01:25.550 UTC" as timestamp) as ts, "C" as id, "EN" as lang
      union all
      select cast("2020-07-03 01:01:25.550 UTC" as timestamp) as ts, "D" as id, "SN" as lang
      union all
      select cast("2020-07-04 01:01:25.550 UTC" as timestamp) as ts, "D" as id, "EN" as lang
    ) where date(ts) between start_dt and end_dt
    group by 1,2

然后输出：

Row     ts_date     id  lang_EN     lang_SN      
1       2020-07-05  A   2           1    
2       2020-07-05  B   2           2    
3       2020-07-05  C   1           1    
4       2020-07-05  D   1           1

【讨论】：

【解决方案2】：

您的查询比必要的要复杂得多。如果您使用COUNT(DISTINCT)，则根本不需要子查询：

SELECT ID, 
       COUNT(DISTINCT Language in ('EN', 'English') THEN DATE(timestamp) END) AS lang_EN,
       COUNT(DISTINCT Language in ('MY', 'Malay') THEN DATE(timestamp) END) AS lang_SN
FROM t
WHERE Smooch_User_ID IS NOT NULL AND
      DATE(Timestamp) <= DATE('2020-01-31') AND
      DATE(Timestamp) >= DATE_ADD(DATE('2020-01-31'), INTERVAL 5 DAY)
GROUP BY ID;

我认为这也捕获了您尝试实现的日期逻辑。

【讨论】：