blocks|key|3429084|text|尝试使用MySQL函数DAYNAME(date1)来检查日期是否为sunday，如果是的话，则得到两天而不是一天的差。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3429085|mysql>+SELECT+DAYNAME(GETDATE());
%2B-----------------------%2B
%7C+DAYNAME('2008-05-15')+%7C
%2B-----------------------%2B
%7C+Thursday++++++++++++++%7C+
%2B-----------------------%2B
1+row+in+set+(0.01+sec)|code-block|syntax|javascript|3429086|如果使用的是SQL+server，则可以尝试datename函数获得日期名称：|3429087|datename(dw,getdate())+|3429088|为Server尝试此操作：|3429089|SELECT+files+from+table1+
where+[date]>=DATEADD(DAY,case+when+datename(dw,(GETDATE()-1))='SUNDAY'+THEN+-2+ELSE+-1+END,getdate());|3429090|对于MySQL服务器，请尝试：|3429091|SELECT+files+from+table1+
where+`date`>=DATE_SUB(current_Date,interval+(case+when+DAYNAME(date_sub(current_date,interval+1+day))='SUNDAY'+THEN+2+ELSE+1+END)+day);|3429092|entityMap^0|B|E|X|6|0|0|M|8|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|18|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|19|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|1A|8|@]|D|@]|E|$]]|$1|U|3|V|5|H|7|1B|8|@]|D|@]|E|$I|J]]|$1|W|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|X|$]]

Try MySQL function <code>DAYNAME(date1)</code> to check for day is <code>sunday</code>,If so then get difference of two days instead of one.

<pre><code>mysql&gt; SELECT DAYNAME(GETDATE());
+-----------------------+
| DAYNAME('2008-05-15') |
+-----------------------+
| Thursday | 
+-----------------------+
1 row in set (0.01 sec)
</code></pre>

If you are using SQL server,you can try <code>datename</code> function to get the day name:

<pre><code>datename(dw,getdate()) 
</code></pre>

Try this for SQL Server:

<pre><code>SELECT files from table1 
where [date]&gt;=DATEADD(DAY,case when datename(dw,(GETDATE()-1))='SUNDAY' THEN -2 ELSE -1 END,getdate());
</code></pre>

For MySQL Server Try:

<pre><code>SELECT files from table1 
where `date`&gt;=DATE_SUB(current_Date,interval (case when DAYNAME(date_sub(current_date,interval 1 day))='SUNDAY' THEN 2 ELSE 1 END) day);
</code></pre>

blocks|key|596957|text|寻找两个日期，当前和之前的日期。从这些记录中拿出新的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|596958|select+files
from+table1+
where+date+=
(+
++select+max(date)+
++from+table1
++where+date+in+(getdate(),+getdate()+-+interval+1+day)
);|code-block|syntax|javascript|596959|(关于守则的更新-Monk的评论。谢谢。)|596960|这是标准的SQL，除了特定于MySQL的日期操作。如果您错误地使用了标记的DBMS，那么将从当前日期减去日期的部分替换为适合您的DBMS的部分。)|596961|更新:您希望从两天中选择的，即currect日和前一天的最后一个日期。由于不应该有未来的记录，这通常意味着表中的两个最大日期。|offset|length|style|BOLD|596962|使用ORDER和altered进行稍微修改的查询：|596963|select+files
from+table1+
where+date+in
(+
++select+date
++from+table1
++order+by+date+desc+limit+2
);|596964|entityMap^0|0|0|0|0|0|D|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@$M|11|N|12|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|T|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

Look for both dates, the current and the one before. From these records take the newer one.

<pre><code>select files
from table1 
where date =
( 
 select max(date) 
 from table1
 where date in (getdate(), getdate() - interval 1 day)
);
</code></pre>

(Updated regarding Code-Monk's commment. Thanks.)

This is standard SQL except for the date manipulation which is MySQL specific. If you were mistaken with the DBMS tagged, then replace the part where a day is subtracted from current day with what's appropriate for your DBMS.)

UPDATE: You want to select from two days, the currect day and the last date before. As there should be no future records, this always means the two maximum dates in your table.

The slightly altered query using ORDER BY and LIMIT:

<pre><code>select files
from table1 
where date in
( 
 select date
 from table1
 order by date desc limit 2
);
</code></pre>

blocks|key|3516464|text|希望这是可行的，因为这里没有mysql来测试(只有Server)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3516465|SELECT+files+FROM+table1
WHERE+CAST([date]+AS+DATE)+IN+(CURDATE()+-+CASE+DATEPART(dw,GETDATE())+WHEN+2+/*+Monday+*/+THEN+2+ELSE+1+END,+CURDATE())|code-block|syntax|javascript|3516466|这样做的目的是测试日期(仅作为日期，而不是时间部分)是否在特定日期列表中，即今天、昨天或周六。|3516467|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Hope this works, because I don't have mysql here to test (only SQL Server):

<pre><code>SELECT files FROM table1
WHERE CAST([date] AS DATE) IN (CURDATE() - CASE DATEPART(dw,GETDATE()) WHEN 2 /* Monday */ THEN 2 ELSE 1 END, CURDATE())
</code></pre>

The idea is to test if the date (as date only, no time part) falls in a specific list of dates, being today and yesterday or Saturday.

blocks|key|596989|text|假设您使用的是Server+(而不是MySQL)，则可以使用交叉应用程序获取文件的最后两个日期的数据：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|596990|SELECT++*
FROM+++dbo.table1+o
+++++++CROSS+APPLY+(+SELECT+TOP+2+[date]
+++++++++++++++++++++FROM+++++dbo.table1+i
+++++++++++++++++++++GROUP+BY+[date]
+++++++++++++++++++++WHERE+[date]>=DATEADD(DAY,DATEDIFF(DAY,0,GETDATE()-5),0)
+++++++++++++++++++++ORDER+BY++[date]+DESC
++++++++++++)+det
WHERE+det.[date]+=+o.[date]|code-block|syntax|javascript|596991|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|$]]

Assuming your are using SQL Server (and not MySQL), you can use a CROSS APPLY to get data for the last two [dates] with files:

<pre><code>SELECT *
FROM dbo.table1 o
 CROSS APPLY ( SELECT TOP 2 [date]
 FROM dbo.table1 i
 GROUP BY [date]
 WHERE [date]&gt;=DATEADD(DAY,DATEDIFF(DAY,0,GETDATE()-5),0)
 ORDER BY [date] DESC
 ) det
WHERE det.[date] = o.[date]
</code></pre>

i need to print the files From the current day plus from the day before...but if the day before is sunday then i want to print the files from saturday

i have a database containing a files of a month..each day produces some files but sunday is a holiday and at that day no files are generated.now i have done the query to pick the files in such a way that giving current date will results the files of that day and previous day. But the problem is on monday when i give the current date it produces files of monday and sunday... i dont want to print files of sunday as it is null,instead of that when am giving the current date if its is monday i should get the files of saturday instead of monday.
this is the code am using now

<pre><code>select files from table1 where [date]&gt;=DATEADD(DAY,DATEDIFF(DAY,0,GETDATE()-1),0) "; 
</code></pre>

except for sundays i have entered the files for each day

<pre><code> insert into dbtable(date,files) values('date value',' filename');
</code></pre>

how to skip a day to previous in sql database if that day produce no files

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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如果前一天是星期天，我需要打印当前日的文件，加上before...but日的文件，然后打印星期六的文件。我有一个数据库，其中包含一个month..each日的文件，生成一些文件，但是星期日是一个假日，在那一天没有文件是generated.now，我已经执行了查询，以这样的方式选择文件，以便给出当前日期将得到当天和前一天...

问如果当天不生成任何文件，如何在sql数据库中跳过一天到前一天。
EN

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问如果当天不生成任何文件，如何在sql数据库中跳过一天到前一天。EN