Ò׽ؽØÍ¼Èí¼þ¡¢µ¥Îļþ¡¢Ãâ°²×°¡¢´¿ÂÌÉ«¡¢½ö160KB

SQLÖÐÂß¼­²éѯ´¦ÀíµÄ¸÷¸ö½×¶Î

 ÓйØSQLÖÐÂß¼­²éѯ´¦ÀíµÄ¸÷¸ö½×¶Î£¨×Ô¼º¸ãµÄÀý×Ó£¬²»¶ÔµÄ»¶Ó­Ö¸ÕýŶ£©
SQL²»Í¬ÓÚÆäËûµÄ±à³ÌÓïÑÔµÄ×î´ó×î´óÌØÕ÷ÓÐ3¸ö°É£¬
Ò»¸öÊÇËüÊÇÃæÏò¼¯ºÏµÄ±à³Ì˼Ï룬µÚ¶þ¸öÊÇÈýÖµÂß¼­£¨Õâ¸öºóÃæ»á˵µ½£©£¬»¹ÓÐÒ»¸ö¾ÍÊǽñÌìÖ÷ҪҪ˵µÄ²éÑ¯ÔªËØµÄÂß¼­´¦Àí´ÎÐò¡£
Çë¿´Ò»¸ö»ù±¾²éѯµÄÂß¼­¹ý³Ì£º
(8)  SELECT (9) DISTINCT (11) <TOP_specification> <select_list>
(1)  from <left_table>
(3)    <join_type> JOIN <right_table>
(2)      ON <join_condition>
(4)  WHERE <where_condition>
(5)  GROUP BY <group_by_list>
(6)  WITH {CUBE | ROLLUP}
(7)  HAVING <having_condition>
(10) ORDER BY <order_by_list>
´ó¼Ò¿ÉÒÔ¿´µ½ ÕâÀïµÄÔËÐв½Öè²»ÊÇÏñÒ»°ãµÄ±à³Ì Ò»¾ä¾ä´ÓÉÏÍùÏ ËüÊÇÌø¶¯µÄ ÓлîÁ¦µÄ
ÕâÀïÌáǰ˵Ï ÿһ²½¶¼»á²úÉúÒ»¸öÐéÄâµÄ±í£¨Ò²¿ÉÄÜÊÇÓα꣬ÏÂÃæ»áÌáµ½£©£¬×÷ΪÏÂÒ»¸ö²½ÖèµÄÊäÈë¡£´ó¼Ò×îºó¿´µ½µÄ½á¹ûÆäʵ¾ÍÊÇ×îºóÒ»¸öÐéÄâ±íÁË¡£
ºÃÁË£¬ÏÂÃæÎÒ¿ªÊ¼¾ßÌå²ûÊÍÿ¸ö²½Ö裺
²âÊÔ»·¾³£º
--ÌâĿҪÇó£ºÇó³öѧÉú×îµÍ¿ÆÄ¿³É¼¨²»µÍÓÚ90·ÖÇÒÄêÁäÔÚ7ËêÒÔÉϵÄѧÉúÐÕÃû
create table #student(s# int,sname varchar(10),age int)
create table #study (s# int, c# char(1),score int)
insert #student
1,'xiaozhu',10 union all select
2,'xiaomao' ,9union all select
3,'xiaozhe' ,7union all select
4,'xiaophai',8 union all select
5,'xiaoduo',9
insert #study select
1,'A',99 union all select
1,'B',90 union all select
1,'C',99 union all select
2,'A',99 union all select
2,'b',99 union all select
2,'c',98 union all select
3,'A',99 union all select
3,'b',92 union all select
3,'c',91 union all select
3,'d',90 union all select
4,'A',88 union all select
4,'B',96
--SQLÓï¾ä
select top 1 sname,MIN(score) as minsocre
from #student s left outer join #study sc
on s.s#=sc.s#
where age>7
group by sname
having MIN(score)>=90
order by minsocre
/*
sname      minsocre
---------- -----------
xiaozhu    90
*/
(1)£ºÖ´Ðеѿ¨¶û»ý£¨CROSS JOIN £©
´ó¼ÒÕâ


Ïà¹ØÎĵµ£º

SQLÓïÑÔ»ù´¡¿¼ºË(Ò»)(oracle)

 1.ÀûÓÃÏÂÃæµÄ½Å±¾´´½¨BOOK£¬READER ºÍ BORROW ±í£¬²¢Íê³ÉºóÃæµÄÁªÏµ¡£
CREATE TABLE BOOK(
    NO CHAR(8) PRIMARY KEY,
    TITLE VARCHAR2(50) NOT NULL,
    AUTHOR VARCHAR2(20) ,
    PUBLISH VARCHAR2(20),
    PUB_DA ......

sql sa怬

 ÎÊÌâÒ»¡¢Íü¼ÇÁ˵ǼMicrosoft SQL Server 2005 µÄsaµÄµÇ¼ÃÜÂë
½â¾ö·½·¨£ºÏÈÓÃwindowsÉí·ÝÑéÖ¤µÄ·½Ê½µÇ¼½øÈ¥£¬È»ºóÔÚ‘°²È«ÐÔ’-‘µÇ¼’-ÓÒ¼üµ¥»÷‘sa’-‘ÊôÐÔ’£¬ÐÞ¸ÄÃÜÂëµã»÷È·¶¨¾Í¿ÉÒÔÁË¡£
ÎÊÌâ¶þ¡¢Òѳɹ¦Óë·þÎñÆ÷½¨Á¢Á¬½Ó£¬µ«ÊÇÔڵǼ¹ý³ÌÖз¢Éú´íÈ¡¡££¨provider:¹ ......

Ò»ÌõSQLÓï¾ä£¬¹ØÓÚ×Ö·û·Ö¸î¹ØÁª¶àÌõ¼Ç¼µÄÎÊÌâ

 Ô­ÎÄ´«ËÍÃÅ£ºhttp://topic.csdn.net/u/20091010/14/FC7737C1-D60B-43F1-A8B5-A9EEF2DE4426.html
¼ÙÈçÏÖÔÚÓÐÁ½ÕÅ±í£º
1.±ístuinfo
sid sname subs
1  jack  |1|2|
2  marry |1|4|
3  tom  |3|
2.±ísubinfo
subid  subname
1      physics
2  &n ......

sybaseÊý¾Ý¿âÖÐË÷Òýµ¼ÖÂsqlÓï¾äÖ´Ðв»³É¹¦

SELECT DISTINCT A.CASEPROP AS PROP,'¾É´æ' AS AJLX,0 AS AJLXXH,A.CASE_PROP AS PROPNO 
 ,M1=( SELECT COUNT(*) from CASES WHERE CASEPROP=A.CASEPROP AND
(PERMITDAY <'2008.12.26 00:00:00' AND (SHUTDAY IS NULL OR
SHUTDAY<'1900-01-01 00:00:00' OR SHUTDAY>='2008.12.26 00:00:00') )) ......
© 2009 ej38.com All Rights Reserved. ¹ØÓÚE½¡ÍøÁªÏµÎÒÃÇ | Õ¾µãµØÍ¼ | ¸ÓICP±¸09004571ºÅ