1. GROUP BY ÊÇ·Ö×é²éѯ, Ò»°ã GROUP BY ÊǺ;ۺϺ¯ÊýÅäºÏʹÓÃ
group by ÓÐÒ»¸öÔÔò,¾ÍÊÇ select ºóÃæµÄËùÓÐÁÐÖÐ,ûÓÐʹÓþۺϺ¯ÊýµÄÁÐ,±ØÐë³öÏÖÔÚ group by ºóÃæ£¨ÖØÒª£©
ÀýÈç,ÓÐÈçÏÂÊý¾Ý¿â±í£º
A B
1 abc
1 bcd
1 asdfg
Èç¹ûÓÐÈçϲéѯÓï¾ä£¨¸ÃÓï¾äÊÇ´íÎóµÄ£¬ÔÒò¼ûÇ°ÃæµÄÔÔò£©
select A,B from table group by A
¸Ã²éѯÓï¾äµÄÒâͼÊÇÏëµÃµ½ÈçϽá¹û(µ±È»Ö»ÊÇÒ»ÏàÇéÔ¸)
A B
abc
1 bcd
asdfg
ÓÒ±ß3ÌõÈçºÎ±ä³ÉÒ»Ìõ,ËùÒÔÐèÒªÓõ½¾ÛºÏº¯Êý£¬ÈçÏÂ(ÏÂÃæÊÇÕýÈ·µÄд·¨):
select A,count(B) as ÊýÁ¿ from table group by A
ÕâÑùµÄ½á¹û¾ÍÊÇ
A ÊýÁ¿
1 3
2. Having
where ×Ó¾äµÄ×÷ÓÃÊÇÔÚ¶Ô²éѯ½á¹û½øÐзÖ×éǰ£¬½«²»·ûºÏwhereÌõ¼þµÄÐÐÈ¥µô£¬¼´ÔÚ·Ö×é֮ǰ¹ýÂËÊý¾Ý£¬Ìõ¼þÖв»Äܰüº¬¾Û×麯Êý£¬Ê¹ÓÃwhereÌõ¼þÏÔÊ¾ÌØ¶¨µÄÐС£
having ×Ó¾äµÄ×÷ÓÃÊÇɸѡÂú×ãÌõ¼þµÄ×飬¼´ÔÚ·Ö×éÖ®ºó¹ýÂËÊý¾Ý£¬Ìõ¼þÖо³£°üº¬¾Û×麯Êý£¬Ê¹ÓÃhaving Ìõ¼þÏÔÊ¾ÌØ¶¨µÄ×飬Ҳ¿ÉÒÔʹÓöà¸ö·Ö×é±ê×¼½øÐзÖ×é¡£
having ×Ӿ䱻ÏÞÖÆ×ÓÒѾÔÚSELECTÓï¾äÖж¨ÒåµÄÁк;ۺϱí´ïʽÉÏ¡£Í¨³££¬ÄãÐèҪͨ¹ýÔÚHAVING×Ó¾äÖÐÖØ¸´¾ÛºÏº¯Êý±í´ïʽÀ´ÒýÓþۺÏÖµ£¬¾ÍÈçÄãÔÚSELECTÓï¾äÖÐ×öµÄÄÇÑù¡£ÀýÈ磺
SELECT A COUNT(B) from TABLE GROUP BY A HAVING COUNT(B)>2
Àý×Óeg£º
ÎÊÌ⣺
Óбí¸ñÈçÏ£º
create table workers(id int primary key,name varchar(12),department varchar(12),salary int,releaseDay date);
Ϊ
id ±àÂë
name ÐÕÃû
department ²¿ÃÅ
salary ¹¤×Ê
releaseDay ·¢·ÅÈÕÆÚ ¸ñʽ 2009-10-10
ÏÖÔÚÒªÇóд³ösqlÓï¾ä£º
ÕÒ³öÔø¾ÔÚÈκÎÒ»¸öÔ·¢ÁËÁ½´Î»òÕßÁ½´Î¹¤×ÊÒÔÉϵÄÔ±¹¤ÐÅÏ¢£¬µ±Ô¹¤×Ê·¢·Å´ÎÊý£¬¹¤×Ê×ÜÊý£¬·¢·ÅÔ·ݣ»
½â´ð£º
create table ty_workers
(
id int primary key,
name varchar(12),
department varchar(12),
salary int,
releaseDay date
);
select * from ty_workers
select name ÐÕÃû, count(salary) ·¢·Å´ÎÊý , sum(salary) ·¢·Å×ܶî, substr(to_char(releaseDay,'yyyy-mm-dd'),0,7) ·¢·ÅÔ·Ý, department ²¿ÃÅ
from ty_workers
gr
ÔÚÆ½Ê±µÄ¹¤×÷¹ý³ÌÖУ¬×÷ΪDBA½ÇÉ«¹ÜÀíÊý¾Ý¿â£¬Í·ÄÔÖеÄÓ¡ÏóÍùÍùÊÇÊý¾Ý¿âʵÀýÃû³Æ£¬¶ø²»»áÈ¥¹ØÐÄServerµÄIP£¬¶ø×÷ΪDeveloperµÄ½ÇÉ«£¬ËûÃÇÍùÍùÏëÖªµÀÖªµÀServer IpºÍ¶Ë¿ÚºÅ¡£ËùÒÔ£¬DBA»á¾³£±»Îʼ°µ½£ºXXXʵÀýµÄIPºÍ¶Ë¿ÚºÅÊÇʲô£¿
Õâ¸öÎÊÌ⣬µ±È»ÎÒÃÇ¿ÉÒÔLoginµ½OS²é¿´IP¡¢Ê¹ÓÃÅäÖÆ¹ÜÀí¹¤¾ß»ñÈ¡µ½¶Ë¿ÚºÅ¡£µ«ÊÇ£¬Õâ¸ö·½·¨·Ç ......