SQLµÄ¼¸¸ö±àÂë¹æ·¶
1.±ÜÃâÔÚwhere×Ó¾äÖжÔ×Ö¶ÎÊ©¼Óº¯Êý£¬ÕâÑù½«µ¼ÖÂË÷ÒýʧЧ£¬±ÈÈ磺
select * from user where
to_char(create_time,'yyyymmdd')='20090101';
ÔÒò£ºÔÚ½¨Á¢indexµÄʱºòÊǸù¾Ý×Ö¶ÎÀ´½¨Á¢µÄ£¬Ò²¾ÍÊÇ˵oracleÔÚinidexµÄʱºòÊÇË÷ÒýµÄ×ֶεÄÖµ£¬Èç¹ûÌṩ¸øoracleµÄÊÇÒ»¸öÐèÒª¾¹ýº¯Êý´¦ÀíµÄ±È½Ï£¬oracle¾Íû°ì·¨Í¨¹ýË÷ÒýÖеÄË÷Òý¼üÖµÀ´½øÐÐÏàÓ¦µÄ±È½Ï£¬ËùÒԾͲ»»á×ßµ½Ë÷ÒýÉÏ
2.±ÜÃâÔÚSQLÖз¢ÉúÒþʽÀàÐÍת»»
È磺
select * from user where id='123';
--ÕâÀïIDÊÇNUMBERÐÍ,»áÔì³Éoracle½«idÏÈת»»³ÉvarcharÀàÐÍÔٱȽϣ¬Ôì³ÉË÷ÒýʧЧ
select * from user where
gmt_create =
to_char('2000-01-01','yyyy-mm-dd');
--ÕâÀïgmt_createÊÇdateÐÍ,»áÔì³Éoracle½«gmt_createÏÈת»»³ÉvarcharÀàÐÍÔٱȽϣ¬Ôì³ÉË÷ÒýʧЧ
3.ȫģºý²éѯÎÞ·¨Ê¹ÓÃINDEX£¬Ó¦µ±¾¡¿ÉÄܱÜÃâ
select * from user where name like '%value%';
4.Èç¹ûʹÓÃOracleÊý¾Ý¿â£¬Ê¹ÓÃOracleµÄÍâÁ¬½Ó£¬¶ø²»ÊDZê×¼µÄÍâÁ¬½ÓÓï·¨
ÕýÈ·£ºselect * from user1 a,user2 b where a.id=b.id(+);
´íÎó£ºselect * from user1 a left join on user2 b a.id=b.id;
5.·ÖÒ³Óï¾ä±ØÐëʹÓÃÈý²ãǶÌ×µÄд·¨
select * from
(select rownum rn,a,* from
(select * from table where Ìõ¼þ
order by Ìõ¼þ) a
where rownum<=100) where rn>80;
Ïà¹ØÎĵµ£º
bcpÊÇSQL ServerÖиºÔðµ¼Èëµ¼³öÊý¾ÝµÄÒ»¸öÃüÁîÐй¤¾ß£¬ËüÊÇ»ùÓÚDB-LibraryµÄ£¬²¢ÇÒÄÜÒÔ²¢Ðеķ½Ê½¸ßЧµØµ¼Èëµ¼³ö´óÅúÁ¿µÄÊý¾Ý¡£bcp¿ÉÒÔ½«Êý¾Ý¿âµÄ±í»òÊÓͼֱ½Óµ¼³ö£¬Ò²ÄÜͨ¹ýSELECT fromÓï¾ä¶Ô±í»òÊÓͼ½øÐйýÂ˺󵼳ö¡£ÔÚµ¼Èëµ¼³öÊý¾Ýʱ£¬¿ÉÒÔʹÓÃĬÈÏÖµ»òÊÇʹÓÃÒ»¸ö¸ñʽÎļþ½«ÎļþÖеÄÊý¾Ýµ¼Èëµ½Êý¾Ý¿â»ò½«Êý¾Ý¿âÖеÄÊý¾Ýµ ......
ת×Ô£ºhttp://tech.ddvip.com/2007-05/117955341625057.html
¼ì²é¸÷Öֱ仯
¡¡¡¡ÎÒÔÚÉè¼ÆÊý¾Ý¿âµÄʱºò»á¿¼Âǵ½ÄÄЩÊý¾Ý×ֶν«À´¿ÉÄܻᷢÉú±ä¸ü¡£±È·½Ëµ£¬ÐÕÊϾÍÊÇÈç´Ë£¨×¢ÒâÊÇÎ÷·½È˵ÄÐÕÊÏ£¬±ÈÈçÅ®ÐÔ½á»éºó´Ó·òÐյȣ©¡£ËùÒÔ£¬ÔÚ½¨Á¢ÏµÍ³´æ´¢¿Í»§ÐÅϢʱ£¬ÎÒÇãÏòÓÚÔÚµ¥¶ÀµÄÒ»¸öÊý¾Ý±íÀï´æ´¢ÐÕÊÏ×ֶΣ¬¶øÇÒ»¹¸½¼ÓÆðʼÈÕºÍÖÕÖ¹ ......
1¡¢SQLÊÇÒ»ÖÖ˵Ã÷ÐÔÓïÑÔ£¬²»Êǹý³Ì»¯ÓïÑÔ¡£ÀàËÆ“¼ìË÷->¼ì²é->²åÈë->¸üДµÄ¹ý³Ì»¯²½ÖèµÄ˳ÐòÊÇûÓÐÒâÒåµÄ¡£Ó¦¸ÃÒÔÐм¯µÄ·½Ê½Ë¼¿¼£¬ÒÔÃèÊöÒ»¸öÂß¼µÄÓïÑÔ·½Ê½Ë¼¿¼¡£
2¡¢ÔÚÉè¼ÆÊý¾Ý¿â±í×Ö¶Îʱ£¬Ò»¸öÐÐÃèÊöÓ¦¸Ã°üº¬Ò»¸öÊÂʵ£¬¶øÇÒÊÇÈ«²¿ÊÂʵ¡£ÀýÈ翼ÇÚ¿¨µÄÉè¼Æ£¬
²»ÒªÉè¼ÆÎª£º[ userId, puchTime, even ......
--> Title : ijÍâÆóSQL ServerÃæ試題
--> Author : wufeng4552
--> Date : 2010-1-15
Question 1£ºCan you use a batch SQL or store procedure to calculating the Number of Days in a Month
Answer 1£ºÕÒ³öµ±ÔµÄÌìÊý
select datepart(dd,dateadd(dd,-1,dateadd(mm,1,cast( ......
²éѯ¼°É¾³ýÖØ¸´¼Ç¼µÄSQLÓï¾ä
1¡¢²éÕÒ±íÖжàÓàµÄÖØ¸´¼Ç¼£¬Öظ´¼Ç¼ÊǸù¾Ýµ¥¸ö×ֶΣ¨peopleId£©À´ÅжÏ
select * from people
where peopleId in (select peopleId from people group by peopleId having count(peopleId) > 1)
2¡¢É¾³ý±íÖжàÓàµÄÖØ¸´¼Ç¼£¬Öظ´¼Ç¼ÊÇ ......