sql – 从to_char()输出中删除空白填充
发布时间:2021-02-24 11:33 所属栏目:116 来源:网络整理
导读:我从这里生成一个视图: create or replace view datetoday asselect to_char(dt,'yyyy-mm-dd') as date,to_char(dt,'Day') as weekday from (select ('2013-03-01'::date + i) dt from generate_series(0,'2013-03-03'::date - 2013-03-01'::date) as t(i))
我从这里生成一个视图: create or replace view datetoday as select to_char(dt,'yyyy-mm-dd') as date,to_char(dt,'Day') as weekday from (select ('2013-03-01'::date + i) dt from generate_series(0,'2013-03-03'::date - 2013-03-01'::date) as t(i)) as t; 它为工作日信息提供了文本类型.然后我用: select date::date,weekday::varchar from datetoday; 现在桌子就像 2013-3-1 Friday 2013-3-2 Saturday 如果我想选择条目: select * from datetoday where weekday='Friday' 将其从文本更改为字符. 似乎长度固定不是根据每个字长. 因为稍后我会将此表的工作日列与另一个表的工作日列进行比较.喜欢 where a.weekday=b.weekday 另一个工作日来自jsp的用户,所以长度不同. 解决方法“Day”模式在空白处填充,使得所有日子都长达9个字符.使用FM Template Pattern Modifier删除任何填充:
SELECT d::date AS day,to_char(d,'yyyy-mm-dd') AS day_text,'FMDay') AS weekday FROM generate_series('2013-03-01'::date,'2013-03-07'::date,interval '1 day') d; 还演示了 (编辑:ASP站长网) |
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