Question

我在SQLite中有以下sql语句：

SELECT
...
CASE WHEN facebook_post_audience >= 10000 THEN 15
WHEN facebook_post_audience >= 3000 AND facebook_post_audience <= 9999 THEN 10
WHEN facebook_post_audience >= 1000 AND facebook_post_audience <= 2999 THEN 7.5
WHEN facebook_post_audience >= 100 AND facebook_post_audience <= 999 THEN 5
ELSE 0 END 'Fb reward category',
facebook_post_number 'No. of FB posts/shares',
'Fb reward category' * facebook_post_number 'FB reward',
...
FROM table;

'FB奖励'应该显示先前定义的列的倍数，但无论这些列中的数字是什么，它都只显示零。此外，typeof（'Fb奖励类别'）表示“文本”，尽管那里只有数字。我今天喝了太多咖啡还是我错过了什么？

Answer 1

将整个案例陈述放入乘法点：

SELECT
...
CASE WHEN facebook_post_audience >= 10000 THEN 15
WHEN facebook_post_audience >= 3000 AND facebook_post_audience <= 9999 THEN 10
WHEN facebook_post_audience >= 1000 AND facebook_post_audience <= 2999 THEN 7.5
WHEN facebook_post_audience >= 100 AND facebook_post_audience <= 999 THEN 5
ELSE 0 END 'Fb reward category',
facebook_post_number 'No. of FB posts/shares',
CASE WHEN facebook_post_audience >= 10000 THEN 15
WHEN facebook_post_audience >= 3000 AND facebook_post_audience <= 9999 THEN 10
WHEN facebook_post_audience >= 1000 AND facebook_post_audience <= 2999 THEN 7.5
WHEN facebook_post_audience >= 100 AND facebook_post_audience <= 999 THEN 5
ELSE 0 END * facebook_post_number 'FB reward',
...
FROM table

Answer 2

使用子查询或CTE。而且，当你在它的时候，给你的列别名给出不需要转义的合理名称：

SELECT ...,
       facebook_post_number,
       fb_reward_category* facebook_post_number as rb_reward,
       ...
FROM (SELECT t.*,
             (CASE WHEN facebook_post_audience >= 10000 THEN 15
                   WHEN facebook_post_audience >= 3000 THEN 10
                   WHEN facebook_post_audience >= 1000 THEN 7.5
                   WHEN facebook_post_audience >= 100 THEN 5
                   ELSE 0
              END) as FB_reward_category 
      FROM table t
     ) t

我还简化了case表达式，因此更容易阅读和维护。

SQLite不会将列相乘

2 个答案: