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group by ÓÐÒ»¸öÔ­Ôò,¾ÍÊÇ select ºóÃæµÄËùÓÐÁÐÖÐ,ûÓÐʹÓþۺϺ¯ÊýµÄÁÐ,±ØÐë³öÏÖÔÚ group by ºóÃæ£¨ÖØÒª£©
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1    bcd
1    asdfg
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select A,B from table group by A

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A     B
       abc
1     bcd
       asdfg 
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select A,count(B) as ÊýÁ¿ from table group by A
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2. Having
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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

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create table workers(id int primary key,name varchar(12),department varchar(12),salary int,releaseDay date);
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id ±àÂë
name ÐÕÃû
department ²¿ÃÅ
salary ¹¤×Ê
releaseDay ·¢·ÅÈÕÆÚ ¸ñʽ 2009-10-10
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id int primary key,
name varchar(12),
department varchar(12),
salary int,
releaseDay date
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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
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