asp排序问题，看似简单,高手来看看，本人一个到就头晕

[diyager]

Dim n
n="a,a,a,b,c,d,d,e,e,e,f,f,g,g"


用asp如何将n里的一串字符按以下排序：

a b c d e f g
a d e f g
a e

[diyager]

楼上的做法像了。

[文盲老顾]

<%
n="a,a,a,b,c,d,d,e,e,e,f,f,g,g"

tmp=split(n,",")
c=split(n,",")

for i=0 to ubound(tmp)-1
c(i)=""
next

for i = 0 to ubound(tmp)-1
j=0:a=0
do while instr(c(j),tmp(i))>0
j=j+1
if j>=ubound(c) then a=1:exit do
loop
if a=1 then
'redim arr(j)
response.end
end if
c(j)=c(j)&" "&tmp(i)
next

for i=0 to ubound(c)
response.write c(i)&"<br>"
if c(i)&" "=" " then exit for
next
%>

[diyager]

唯一的要求就是
a或者b或者c，只要重复次数超过5次，就另起一列

[文盲老顾]

没看到规则变了 -_-b

[小码]

<%'更正：
function a(byval b)
aa=""
m = split (b,",")
for i=0 to Ubound(m)-1
if m(i)=m(i+1) then
aa=aa+m(i)&","
else
If i>0 Then
j=asc(m(i))-asc(m(i-1))
If j>1 Then
For k=1 to j-1
response.write "-"
Next
End If
End If
response.write m(i)
end if
next
response.write "<br>"
if aa<>"" then
a(aa)
end if
end function
a("a,a,a,b,c,d,d,e,e,e,f,f,g,g,")
%>
有一点不明白，重复的留作下次再排，怎么会有超过2次甚至达到5次的时候呢？请把最终结果例举出来看看？

[gq]

<%
dim j
dim a1,a2,a3,a4,a5
j=1
function a(b)
m = split (b,",")
for i=0 to Ubound(m)-1
if m(i)= m(i+1) then
if j=6 then
j=1
end if
select case j
case 1
a1=a1+m(i) &","
case 2
a2=a2+m(i)&","
case 3
a3=a3+m(i)&","
case 4
a4=a4+m(i)&","
case 5
a5=a5+m(i)&","
end select
j = j+1
else
select case j
case 1
a1=a1+m(i)&","
case 2
a2=a2+m(i)&","
case 3
a3=a3+m(i)&","
case 4
a4=a4+m(i)&","
case 5
a5=a5+m(i)&","
end select
j=1
end if
next
end function
a("a,a,a,a,a,a,a,a,a,b,b,c,d,d,e,e,e,f,f,g,g,g,")
response.write a1 &"<br>"
response.write a2 &"<br>"
response.write a3 &"<br>"
response.write a4 &"<br>"
response.write a5 &"<br>"

m1 = split (a1,",")
m2 = split (a2,",")
response.write replace(a1,",","") &"<br>"
g=0
for i = 0 to Ubound(m1)-1
g=0
for j = 0 to Ubound(m2)-1
if m1(i)=m2(j) then
g = 1
end if
next
if g=1 then
response.write m1(i)
else
response.write "-"
end if
next
response.write "<br>"

m3 = split (a3,",")
g=0
for i = 0 to Ubound(m1)-1
g=0
for j = 0 to Ubound(m3)-1
if m1(i)=m3(j) then
g = 1
end if
next
if g=1 then
response.write m1(i)
else
response.write "-"
end if
next
response.write "<br>"
m4 = split (a4,",")
g=0
for i = 0 to Ubound(m1)-1
g=0
for j = 0 to Ubound(m4)-1
if m1(i)=m4(j) then
g = 1
end if
next
if g=1 then
response.write m1(i)
else
response.write "-"
end if
next
response.write "<br>"
m5 = split (a5,",")
g=0
for i = 0 to Ubound(m1)-1
g=0
for j = 0 to Ubound(m5)-1
if m1(i)=m5(j) then
g = 1
end if
next
if g=1 then
response.write m1(i)
else
response.write "-"
end if
next

%>

[gq]

纳闷。。真是无聊了我。

[dong127]

写点程序不是正好打消了你的无聊吗

[net205]

咋有这样的要求。。。。

看热闹.

[taochunsong]

我智商太低了
没有看懂
到底什么规律阿

[xiaoyaowp]

按顺序抽字母（从每次剩下的）
循环

[dong127]

回家后再给个我写的

[三仙半]

这个帖可以写个标题叫“智力测验”了，好帖，坚决建议斑竹给加分！

[diyager]

<%
Dim n
dim s,t,d
n="a,a,a,b,c,d,e,r,f,d,e,g,g,g,f,a,a"
arrn=split(n,",")
arrnlen=Ubound(arrn)

dim charnew,charold,j,k
redim arrNew(arrnlen,arrnlen)
j=0
k=0
charnew=""
charold=""

for i=0 to ubound(arrn)
charnew=arrn(i)
if charnew=charold then
j=j+1
if j>maxJ then maxJ=j
else
j=1
k=k+1
end if
arrNew(j,k)=charnew
charold=charnew
next
maxK=k

response.write "原字符串："&n&"<br>"
response.write "转换后字符串：<br>"
for j=0 to maxJ
for k=0 to maxK
if len(arrNew(j,k))>0 then
response.write arrNew(j,k)&" "
else
response.write "　"
end if
next
response.write "<br>"
next
%>

[diyager]

已经解决。谢谢！

[dong127]

解决了？我怎么测试出来不对的？

输出：
原字符串：a,a,a,a,a,a,a,a,a,a,a,b,c,d,e,r,f,d,e,g,g,g,f,f,f,f,f,f,f,f,f,f,f,f,f,f,f,f
转换后字符串：
　　　　　　　　　　　　
　a b c d e r f d e g f
　a 　　　　　　　　g f
　a 　　　　　　　　g f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　a 　　　　　　　　　f
　　　　　　　　　　　f
　　　　　　　　　　　f
　　　　　　　　　　　f
　　　　　　　　　　　f
　　　　　　　　　　　f

[文盲老顾]

^v^，我也没搞明白

谢谢加分 ^v^

[xuanstar]

先拆分，再分析结果：第三组数据只出现三次，第二组数据都出现了两次，第一组就是 distinct 所有的数据了。

[xuanstar]

将字符串传人sql，再拆分，从出现最多次数的开始循环，就可以了。在存储过程解决这个问题，比较方便。

[超级大笨狼]

白痴，这个根本不需要排序，只要找出多少个a,多少个b放到表格里就行了。