P
popzhu
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SELECT COUNT(DISTINCT *) FROM TABLENAME?
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gophie
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关注
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SuperJS
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select count * from tablename groupby colname
哈
哈利波特
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关注。
U
ugvanxk
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select count(*) from
(
select allfieldname,count(*) from tablename
having count(*)=1
group by allfieldname
)
(
select allfieldname,count(*) from tablename
having count(*)=1
group by allfieldname
)
P
popzhu
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我的问题可能不是太清楚,说明一下
SELECT ZIP,COUNT(DISTINCT NAME) FROM TABLENAME
GROUP BY ZIP
要求在这个邮编下,所有不同名字的人的个数!
SELECT ZIP,COUNT(DISTINCT NAME) FROM TABLENAME
GROUP BY ZIP
要求在这个邮编下,所有不同名字的人的个数!
U
ugvanxk
Unregistered / Unconfirmed
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select zip, count(*) from
(
select zip,name from tablename
having count(*)=1
group by zip,name
)
group by zip
(
select zip,name from tablename
having count(*)=1
group by zip,name
)
group by zip
P
popzhu
Unregistered / Unconfirmed
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ugvanxk你的答案不正确
1, 如果记录重复的NAME,你的方法是统计不进去的,你ONLY统计一条记录的NAME
2/ HAVING COUNT(*) 应该在GROUP BY 后面吧!
1, 如果记录重复的NAME,你的方法是统计不进去的,你ONLY统计一条记录的NAME
2/ HAVING COUNT(*) 应该在GROUP BY 后面吧!
P
popzhu
Unregistered / Unconfirmed
GUEST, unregistred user!
有没有人知道!
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gophie
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select zip, count(*) from
(
select zip,name from tablename
group by zip,name
)
group by zip
可以是可以，
但ACCESS没有类似UNIQUE这样的函数？
(
select zip,name from tablename
group by zip,name
)
group by zip
可以是可以，
但ACCESS没有类似UNIQUE这样的函数？
Z
zhanzehua
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select zip,name,count(*) from tablename
group by zip,name
这样难道不行
group by zip,name
这样难道不行
P
popzhu
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不行
这样是同一个邮编同一个户名下有多少人家,
并非同一个邮编下,有多少个不同的人家!
这样是同一个邮编同一个户名下有多少人家,
并非同一个邮编下,有多少个不同的人家!
Y
yaoluo
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select zip, count(*) from (select distinct (zip+name),zip from tablename) group by zip
Z
zhanzehua
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可不可以用中间表
select zip,name into TempTable from tablename
group by zip,name
select zip,count(*) from TempTable
group by zip
select zip,name into TempTable from tablename
group by zip,name
select zip,count(*) from TempTable
group by zip
U
ugvanxk
Unregistered / Unconfirmed
GUEST, unregistred user!
select zip, count(*) from
(
select distinct zip,name //name,zip
from tablename
)
group by zip
我的应该能通过，在access 试了一下
elect kind,count(*) from
(select distinct alpha,kind from tbl_fix_time)
group by kind
(
select distinct zip,name //name,zip
from tablename
)
group by zip
我的应该能通过，在access 试了一下
elect kind,count(*) from
(select distinct alpha,kind from tbl_fix_time)
group by kind
P
popzhu
Unregistered / Unconfirmed
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我希望不用嵌套语句,直接有类是 count(distinct name)
的效果!
的效果!
P
popzhu
Unregistered / Unconfirmed
GUEST, unregistred user!
多人接受答案了。