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id name kecheng chengji
1 ÕÅÈý ÓïÎÄ 70
2 ÕÅÈý Êýѧ 90
3 ÀîËÄ ÓïÎÄ 85
4 ÀîËÄ Êýѧ 87
5 ÀîËÄ Ó¢Óï 88
6 ÍõÎå ÓïÎÄ 79
7 ÍõÎå Êýѧ 90
8 ÍõÎå Ó¢Óï 82


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select name, count(chengji) as ×Ü·Ö from ±íÃû where ×Ü·Ö>80 group by name


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SQL code:
select name from tablename where sum(chengji)>80 group by name


select [name] from biao group by [name] having (sum(chengji)/count([name])) > 80

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select name from [tableName] where chengji > 80 group by name having count(*) >= 3

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select [name] from (select [name], min(chengji) min


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