话说某日去面试，自己脑残很简单的sql都写不好，得多补补基础了，平日里总眼高手低可不行。
记录下来，是个教训.....
题目很简单，有一test[id,name,age]表，结构如下：
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1.查出name中含a的记录：select * from test t where t.name like '%a%'
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2.查出同一年龄段人的数目：select age,count(*) from test group by age
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3.查出含有相同名称的人的名字：select? name from test group by name having count(name) > 1
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4.查具有相同名字人的记录：select * from test where name in (select name from test group by name having count(name) > 1)