superset如何按照日期数量聚合查询

需要写一个 sql 从表 ods.ods_cc_ad_log 查询 dt 范围是 ‘2023-09-01’ 到 ‘2023-09-08’ 每一天的creative_id 的总数,麻烦给出具体的 sql 语句: 以下是你需要的SQL语句: SELECT DATE(created_time) AS dt, COUNT(DISTINCT creative_id) AS creative_id_count FROM ods.ods_cc_ad_log WHERE DATE(created_time) BETWEEN ‘2023-09-01’ AND ‘2023-09-08’ GROUP BY 1 ORDER BY 1;

这个SQL语句做了以下操作: FROM子句从ods.ods_cc_dsp_log表查询数据 WHERE子句过滤created_time字段处于’2023-09-01’到’2023-09-08’之间的记录 SELECT子句选择created_time日期字段(用DATE函数提取日期部分)和creative_id字段进行分组计数后的值 GROUP BY子句按日期分组 ORDER BY子句按日期升序排序结果 这样就可以查询指定日期范围内每一天creative_id的总数了 下面是有关 SQL 查询的一些其他详细信息: DATE(created_time) 仅从created_time列中提取日期部分,因此我们可以按日期而不是按完整日期时间进行分组和计数。 COUNT(非重复creative_id)仅统计每个日期组中的唯一/不同creative_id值,而不是所有行。这样可以准确统计每天看到的不同广告素材 ID 的数量。 在“2023-09-01”和“2023-09-08”之间,将行筛选为指定的 8 天范围。开始日期和结束日期都包括在内。 GROUP BY 1 按所选的第一列(即 DATE(created_time) 别名)对结果进行分组。我们也可以显式使用 GROUP BY DATE(created_time)。 ORDER BY 1 按升序日期顺序对结果进行排序,因此输出按逻辑时间顺序排列。 这将返回 8 行,指定范围内的每个日期对应一行,其中包含该日期的日期和广告素材 ID 计数。 像 COUNT 这样的聚合函数会忽略 NULL 值。因此,任何缺少creative_id的行都不会影响计数。 created_time和creative_id上的索引可以通过更有效地启用筛选和聚合来提高大型表的性能。 将日期格式化为“YYYY-MM-DD”可确保它正确排序为日期而不是字符串。 如果查询结构或逻辑的任何部分需要更多解释,请告诉我! 你能更详细地解释一下COUNT(DISTINCT creative_id)函数是如何工作的吗? Claude-instant: 当然,这里有一个关于COUNT(DISTINCT)如何工作的更详细的解释: COUNT 通常只计算每组中的总行数。这包括重复值。 COUNT(DISTINCT) 改为仅计算表达式内部的唯一/不同值。 它的工作原理是首先识别每个日期组中creative_id的所有唯一值。 然后计算有多少个唯一值。 例如,如果在 2023-09-01 上有以下行: 广告素材 ID 1 1 2 3 COUNT 将返回 4,因为它对所有行进行计数 COUNT(非重复creative_id)将首先看到唯一值为1,2,3 然后它将返回 3,因为有 3 个不同/唯一的广告素材 ID 这样可以准确统计不同的广告素材 ID 的数量,而不是包含重复项的总行数。 数据库通过内部执行排序、散列或索引等操作来评估 COUNT(DISTINCT),以有效地识别组中的唯一值。 因此,它避免了执行昂贵的 DISTINCT 查询,然后是聚合,并在“后台”执行所有查询以获得最佳性能。 让我知道 COUNT(DISTINCT) 如何识别和计算组内唯一值的任何部分是否需要更多解释! Here are some additional details about the SQL query: DATE(created_time) extracts just the date part from the created_time column, so we can group and count by date rather than by full datetime. COUNT(DISTINCT creative_id) counts only unique/distinct creative_id values within each date group, rather than all rows. This gives an accurate count of how many different creative IDs were seen each day. BETWEEN ‘2023-09-01’ AND ‘2023-09-08’ filters the rows to just the 8 day range specified. Both the start and end dates are inclusive. GROUP BY 1 groups the results by the first column selected, which is the DATE(created_time) alias. We could also explicitly use GROUP BY DATE(created_time). ORDER BY 1 sorts the results in ascending date order, so the output is in logical chronological sequence. This would return 8 rows, one for each date in the specified range, with the date and creative ID count for that date. Aggregate functions like COUNT ignore NULL values. So any rows missing the creative_id would not affect the counts. Indexes on created_time and creative_id could improve performance for large tables by enabling filtering and aggregation more efficiently. Formatting the date as ‘YYYY-MM-DD’ ensures it properly sorts as a date rather than a string. Can you explain how the COUNT(DISTINCT creative_id) function works in more detail? Claude-instant-100k: Sure, here’s a more detailed explanation of how COUNT(DISTINCT) works: COUNT normally just counts the total number of rows in each group. This includes duplicate values. COUNT(DISTINCT) instead counts only unique/distinct values of the expression inside. It works by first identifying all unique values of creative_id within each date group. Then it counts how many unique values there are. For example, if on 2023-09-01 there were rows with: Creative ID 1 1 2 3 COUNT would return 4, since it counts all rows COUNT(DISTINCT creative_id) would first see the unique values are 1, 2, 3 It would then return 3, since there are 3 distinct/unique creative IDs This gives an accurate count of how many different creative IDs, rather than total rows which could include duplicates. The database evaluates COUNT(DISTINCT) by internally performing operations like sorting, hashing or indexing to efficiently identify unique values within groups. So it avoids doing costly DISTINCT queries followed by aggregate, and performs it all “under the hood” for best performance. Let me know if any part of how COUNT(DISTINCT) identifies and counts unique values within groups needs more explanation!

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