Question

我有一张看起来像下面的桌子

| id | views                                                                     | actions                                                                   |
|----|---------------------------------------------------------------------------|---------------------------------------------------------------------------|
| 1  | {"group1":{"mobile":20,"desktop":36},"group2":{"mobile":25,"desktop":41}} | {"group1":{"mobile":4,"desktop":3},"group2":{"mobile":2,"desktop":4}}     |
| 2  | {"group1":{"mobile":40,"desktop":34},"group2":{"mobile":31,"desktop":45}} | {"group1":{"mobile":21,"desktop":19},"group2":{"mobile":23,"desktop":19}} |
| 3  | {"group1":{"mobile":53,"desktop":34},"group2":{"mobile":37,"desktop":75}} | {"group1":{"mobile":17,"desktop":14},"group2":{"mobile":30,"desktop":12}} |

我希望我的结果看起来像下面的

| product     | views group 1  | views group 2  | actions group 1 | actions group 2 | actions per view group 1 | actions per view group 2 |
|-------------|----------------|----------------|-----------------|-----------------|--------------------------|--------------------------|
| 1 (mobile)  | 20             | 25             | 4               | 2               | (D2/B2) = 0.2            | (E2/C2) = 0.08           |
| 1 (desktop) | 36             | 41             | 3               | 4               | (D3/B3) = 0.115          | (E2/C2) = 0.09           |
| 2 (mobile)  | 40             | 31             | 21              | 23              | ...                      | ...                      |
| 2 (desktop) | 34             | 45             | 19              | 19              | ...                      | ...                      |
| 1 (overall) | (B2 + B3) = 56 | (C2 + C3) = 66 | ...             | ...             | ...                      | ...                      |
| 2 (overall) | (B4 + B5) = 84 | (C4 + C5) = 76 | ...             | ...             | ...                      | ...                      |

我可以使用以下SQL查询来执行此操作，但是我想知道是否可以以更简洁或更短的方式进行操作

SELECT
    CASE id WHEN 1 THEN '1 (mobile)' WHEN 2 THEN '2 (mobile)' END as product,
    SUM(json_extract_path_text(actions, 'group1', 'mobile') AS actions_group_1,
    SUM(json_extract_path_text(views, 'group1', 'mobile') AS views_group_1,
    SUM(json_extract_path_text(actions, 'group2', 'mobile') AS actions_group_2,
    SUM(json_extract_path_text(views, 'group2', 'mobile') AS views_group_2,
    SUM(json_extract_path_text(actions, 'group1', 'mobile')/SUM(json_extract_path_text(views, 'group1', 'mobile') AS actions_per_view_group_1 ,
    SUM(json_extract_path_text(actions, 'group2', 'mobile')/SUM(json_extract_path_text(views, 'group2', 'mobile') AS actions_per_view_group_2
FROM widget_daily_stats
WHERE id IN (1,2)
GROUP BY id

UNION

SELECT
    CASE id WHEN 1 THEN '1 (desktop)' WHEN 2 THEN '2 (desktop)' END as product,
    SUM(json_extract_path_text(actions, 'group1', 'desktop') AS actions_group_1,
    SUM(json_extract_path_text(views, 'group1', 'desktop') AS views_group_1,
    SUM(json_extract_path_text(actions, 'group2', 'desktop') AS actions_group_2,
    SUM(json_extract_path_text(views, 'group2', 'desktop') AS views_group_2,
    SUM(json_extract_path_text(actions, 'group1', 'desktop')/SUM(json_extract_path_text(views, 'group1', 'desktop') AS actions_per_view_group_1 ,
    SUM(json_extract_path_text(actions, 'group2', 'desktop')/SUM(json_extract_path_text(views, 'group2', 'desktop') AS actions_per_view_group_2
FROM widget_daily_stats
WHERE id IN (1,2)
GROUP BY id

UNION

...

我必须有一种方法可以遍历移动设备和台式机并动态地统一结果，而不必显式地统一所有结果

Answer 1

首选解决方案是不要在Amazon Redshift中将数据存储为JSON 。如您所见，您需要使用json_extract_path_text()函数来检索单个元素，这很慢。

此外，将字段存储在JSON中意味着Redshift无法优化存储和访问。如果数据存储在单独的字段中（如您在输出中显示的那样），则Redshift效率会更高。

因此：

尝试将数据直接加载到字段中，而不是作为JSON对象加载（使用COPY from JSON Format）
或者，将JSON数据加载到登台表中，然后使用如上所写的查询来提取字段，然后将其插入到新表中。然后使用该新表进行所有查询。它们将更高效地运行。

很抱歉，此答案并未专门解决您的UNION问题，但您会发现将数据转换为更可用的格式将有助于您查询数据。

使用UNION动态循环/加入结果

1 个答案: