我没有时间戳表。我只需要使用名称字符串进行转换,
SELECT day_of_week;
输出应类似于
day_of_week sum
Friday 5
Monday 1
Saturday 6
Sunday 7
Thursday 4
Tuesday 2
Wednesday 3
答案 0 :(得分:0)
如果是时间戳,则可以使用isodow
。参见doc
isodow
星期几为星期一(1)至星期日(7)
SELECT day_of_week, EXTRACT(ISODOW FROM TIMESTAMP '2001-02-18 20:38:40') as sum FROM your_table;
但是您没有该列。因此,您可以像下面一样使用CASE...WHEN
-
SELECT day_of_week,
CASE WHEN day_of_week='Monday' THEN 1
WHEN day_of_week='Tuesday' THEN 2
WHEN day_of_week='Wednesday' THEN 3
WHEN day_of_week='Thursday' THEN 4
WHEN day_of_week='Friday' THEN 5
WHEN day_of_week='Saturday' THEN 6
ELSE 7
END as sum
FROM your_table;
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