一个比较复杂的查询语句
2010-07-13 16:09
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Access中不支持视图,认为视图是个复杂查询。
Create View TV as
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
union
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="暂不推荐" group by B.IPDCode,B.Grade;
Access中不认识关键字“View”。
第一版查询语句:
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
union
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="暂不推荐" group by B.IPDCode,B.Grade;
第二版改进型查询语句:
select B.IPDCode as 类别,count(B.Grade) as 建议推荐,
(select count(B.Grade) From WW_IPD C,WW_SM D Where C.IPDCode = D.IPDCode And D.Grade="暂不推荐" AND C.IPDCode=B.IPDCode )as 暂不推荐
From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
第二版语句有漏洞,下面为第三版改进型查询语句:
select B.IPDCode as 类别,
(select count(B.Grade) From WW_IPD C,WW_SM D Where C.IPDCode = D.IPDCode And D.Grade="暂不推荐" AND C.IPDCode=B.IPDCode )as 暂不推荐, count(B.Grade) as 建议推荐
From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
union
select B.IPDCode as 类别,count(B.Grade) as 暂不推荐,
(select count(B.Grade) From WW_IPD C,WW_SM D Where C.IPDCode = D.IPDCode And D.Grade="建议推荐" AND C.IPDCode=B.IPDCode )as 建议推荐
From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="暂不推荐" group by B.IPDCode,B.Grade
*感悟:用户需求促使我们技术探索!
Create View TV as
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
union
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="暂不推荐" group by B.IPDCode,B.Grade;
Access中不认识关键字“View”。
第一版查询语句:
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
union
select B.IPDCode as 类别,B.Grade,count(B.Grade) as 数目 From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="暂不推荐" group by B.IPDCode,B.Grade;
第二版改进型查询语句:
select B.IPDCode as 类别,count(B.Grade) as 建议推荐,
(select count(B.Grade) From WW_IPD C,WW_SM D Where C.IPDCode = D.IPDCode And D.Grade="暂不推荐" AND C.IPDCode=B.IPDCode )as 暂不推荐
From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
第二版语句有漏洞,下面为第三版改进型查询语句:
select B.IPDCode as 类别,
(select count(B.Grade) From WW_IPD C,WW_SM D Where C.IPDCode = D.IPDCode And D.Grade="暂不推荐" AND C.IPDCode=B.IPDCode )as 暂不推荐, count(B.Grade) as 建议推荐
From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="建议推荐" group by B.IPDCode,B.Grade
union
select B.IPDCode as 类别,count(B.Grade) as 暂不推荐,
(select count(B.Grade) From WW_IPD C,WW_SM D Where C.IPDCode = D.IPDCode And D.Grade="建议推荐" AND C.IPDCode=B.IPDCode )as 建议推荐
From WW_IPD A,WW_SM B Where A.IPDCode = B.IPDCode And B.Grade="暂不推荐" group by B.IPDCode,B.Grade
*感悟:用户需求促使我们技术探索!
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