blocks|key|2543335|text|如果未指定日期的时间部分，则为00:00:00。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2543336|如果你担心时间部分，你总是可以截断时间部分：|2543337|Trunc(to_date('16.07.2010',+'dd.MM.YYYY'))+and+
Trunc(to_date('20.07.2010',+'dd.MM.YYYY'))|code-block|syntax|javascript|2543338|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you does not specify time part of date it will be 00:00:00.

If you worry about time part you always can truncate time part:

<pre><code>Trunc(to_date('16.07.2010', 'dd.MM.YYYY')) and 
Trunc(to_date('20.07.2010', 'dd.MM.YYYY'))
</code></pre>

blocks|key|790106|text|最简单的方法是使用ANSI日期文字，它不允许指定时间，因此在内部使用00:00:00时间部分：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|790107|dateadded+between+date+'2010-07-16'+and+date+'2010-07-20'|code-block|syntax|javascript|790108|您的表达式看起来很奇怪，因为您的日期字符串包含点作为时间分量分隔符，但是您的日期格式包含逗号。但是Oracle接受它：|790109|rwijk@XE>+select+to_date('16.07.2010',+'dd,MM,YYYY')+from+dual;

TO_DATE('16.07.2010
-------------------
16-07-2010+00:00:00

1+row+selected.|790110|为了简单起见，我使用ANSI日期文字。|790111|问候你，罗伯。|790112|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The easiest way is to use ANSI date literals, which doesn't allow time to be specified and is thus with a 00:00:00 time part internally:

<pre><code>dateadded between date '2010-07-16' and date '2010-07-20'
</code></pre>

Your expressions look odd, as your date string has dots as the time component separator, but your date format has commas. Oracle accepts it though:

<pre><code>rwijk@XE&gt; select to_date('16.07.2010', 'dd,MM,YYYY') from dual;

TO_DATE('16.07.2010
-------------------
16-07-2010 00:00:00

1 row selected.
</code></pre>

I'd use the ANSI date literal for its simplicity.

Regards,
Rob.

blocks|key|2543405|text|此查询将返回一行，回答您的问题我认为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2543406|SELECT+TO_DATE('16.07.2010+00:00:00',+'dd,MM,YYYY+HH24:Mi:SS'),++
+++++++TO_DATE('16.07.2010',+'dd,MM,YYYY')
++FROM+dual
+WHERE+to_date('16.07.2010+00:00:00',+'dd,MM,YYYY+HH24:Mi:SS')+=+to_date('16.07.2010',+'dd,MM,YYYY')++|code-block|syntax|javascript|2543407|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

this query will returns you one row which answers your question i think :

<pre><code>SELECT TO_DATE('16.07.2010 00:00:00', 'dd,MM,YYYY HH24:Mi:SS'), 
 TO_DATE('16.07.2010', 'dd,MM,YYYY')
 FROM dual
 WHERE to_date('16.07.2010 00:00:00', 'dd,MM,YYYY HH24:Mi:SS') = to_date('16.07.2010', 'dd,MM,YYYY') 
</code></pre>

blocks|key|1536498|text|这是SQL一以贯之的一件事--如果您不在包含时间部分的DATE数据类型上提供时间部分(IE+Oracle、PostgreSQL，而MySQL和SQL+Server显式地将其称为DATETIME)，那么该时间部分将被默认为该日期的午夜-+00:00:00。如果是午夜，这意味着日期的开始--如果您想在日期比较中包含整个日期，则该值的时间部分需要是23:49:49+--午夜前一秒。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1536499|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

This is one thing that SQL is consistent about - if you don't provide a time portion on a DATE data type that includes a time portion (IE Oracle, PostgreSQL while MySQL and SQL Server explicitly call it DATETIME), then the time portion is defaulting to exactly midnight of that date - 00:00:00. And being midnight, that means the start of the date - if you want to include the entire date in a date comparison, the value's time portion needs to be 23:49:49 -- a second to midnight.

blocks|key|790174|text|23:59:59离午夜还有一秒。本质上，午夜被定义为一天的开始:+00:00:00。然后秒表计时到一天结束。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|790175|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

23:59:59 is a second to midnight. Essentially midnight is defined as the start of a day: 00:00:00. Then the seconds clock up to the end of the day.

Heyho,

I need to grab some datas from actions which been done from date A 00:00:00 and date B 00:00:00 (in this case
Date A: 16.07.2010
Date B: 20.07.2010)
so i wrote this select-statement:

<pre><code>Select avg(cnt),fext from (
 Select 
 to_char(mytable.dateadded, 'DD.MM.YYYY') dateadded,
 fext, 
 count(id) cnt 
 from mytable 
 where dateadded between
 to_date('16.07.2010', 'dd,MM,YYYY') and 
 to_date('20.07.2010', 'dd,MM,YYYY')
 group by 
 to_char(mytable.dateadded, 'DD.MM.YYYY'),
 fext)
group by fext;
</code></pre>

The original (and working) statement had:

<pre><code> to_date('16.07.2010 00:00:00', 'dd,MM,YYYY HH24:Mi:SS') and 
 to_date('20.07.2010 00:00:00', 'dd,MM,YYYY HH24:Mi:SS')
</code></pre>

so the question is: does the 

<pre><code> to_date('16.07.2010', 'dd,MM,YYYY') and 
 to_date('20.07.2010', 'dd,MM,YYYY')
</code></pre>

already set the time to date A and B to 00:00:00?

Greetz

Is a date in Oracle-SQL without a time-value always 00:00:00?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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嘿，嘿，我需要从日期A 00:00:00和日期B 00:00:00 (在本例中日期A: 16.07.2010，日期B: 20.07.2010)执行的操作中获取一些数据，因此我编写了以下select语句：Select avg(cnt),fext from (        Select             to_ch...

问Oracle-SQL中没有时间值的日期总是00:00:00吗？
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问Oracle-SQL中没有时间值的日期总是00:00:00吗？EN