SQL 甲骨文日期范围
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oracle date range
提问by Hoax
using a Oracle 10g db I have a table something like this:
使用 Oracle 10g db 我有一个像这样的表：
create table x(
ID NUMBER(10) primary key,
wedding DATE NOT NULL
);
how can I
我怎样才能
select * from x where wedding is in june 2008???
I know it is probably an easy one but I couldn't find any satisfying answer so far. Help is very much appreciated.
我知道这可能很简单，但到目前为止我找不到任何令人满意的答案。非常感谢帮助。
回答by OMG Ponies
Use:
用：
SELECT *
FROM x
WHERE x.wedding BETWEEN TO_DATE('2008-JUN-01', 'YYYY-MON-DD')
AND TO_DATE('2008-JUL-01', 'YYYY-MON-DD')
Use of TO_DATEconstructs a date with a time portion of 00:00:00, which requires the end date to be one day ahead unless you want to use logic to correct the current date to be one second before midnight. Untested:
Use ofTO_DATE构造一个时间部分为 00:00:00 的日期，这要求结束日期提前一天，除非您想使用逻辑将当前日期更正为午夜前一秒。未经测试：
TO_DATE('2008-JUN-30', 'YYYY-MON-DD') + 1 - (1/(24*60*60))
That should add one day to 30-Jun-2008, and then subtract one second in order to return a final date of 30-Jun-2008 23:59.
这应该将 30-Jun-2008 加一天，然后减去一秒以返回30-Jun-2008 23:59.
References:
参考：
回答by Roland Bouman
This is ANSI SQL, and supported by oracle as of version 9i
这是 ANSI SQL，从版本 9i 开始由 oracle 支持
SELECT *
FROM x
WHERE EXTRACT(YEAR FROM wedding) = 2008
AND EXTRACT(MONTH FROM wedding) = 06
Classic solution with oracle specific TO_CHAR():
oracle特定的经典解决方案TO_CHAR()：
SELECT *
FROM x
WHERE TO_CHAR(wedding, 'YYYY-MMM') = '2008-JUN'
(the latter solutions was supported when dinosaurs still walked the earth)
（当恐龙还在地球上行走时，支持后一种解决方案）
