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wren135 2005-04-04
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顶楼上的
qiqunet 2005-04-04
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function splitNum(a)
for i=1 to len(a)
if not IsNumeric(right(a,i)) then
splitNum=right(a,i-1)
exit for
end if
next
end function
msgbox(splitNum("123aaa-321"))
for i=1 to len(a)
if not IsNumeric(right(a,i)) then
splitNum=right(a,i-1)
exit for
end if
next
end function
msgbox(splitNum("123aaa-321"))
Hozaka 2005-04-04
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CInt(Right("123abc12", 2))
结果就是 12 ，数字型的
结果就是 12 ，数字型的
fangq 2005-04-04
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规则的直接用right(文本，２)
lookatliu 2005-04-04
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用mid函数啦
datamoon 2005-04-04
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对，规则的。
hbhbhbhbhb1021 2005-04-04
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<%
dim regEx,Matches
dim str1
str1="123asd12 "
set regEx=new regExp
regEx.Pattern= "([0123456789.]+?)([abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ]+?)([0123456789.]+?)(\s)"
regEx.IgnoreCase=true
regEx.Global=True
set Matches=regEx.Execute(str1)
for each Match in Matches
Response.Write Match.Submatches(2) & "<br>"
next
set Matches=nothing
set regEx=nothing
%>
dim regEx,Matches
dim str1
str1="123asd12 "
set regEx=new regExp
regEx.Pattern= "([0123456789.]+?)([abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ]+?)([0123456789.]+?)(\s)"
regEx.IgnoreCase=true
regEx.Global=True
set Matches=regEx.Execute(str1)
for each Match in Matches
Response.Write Match.Submatches(2) & "<br>"
next
set Matches=nothing
set regEx=nothing
%>
wren135 2005-04-04
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所有串的是有规则的吗
如152aaa15
231sdf55
213efs89
的吗
如152aaa15
231sdf55
213efs89
的吗
