如何判断电子邮件的地址格式是否正确？

答案：
    我们可用下列办法来解决这一问题——但只是能够判断每个电子邮件地址的格式是否有效，并不能保证该地址确实存在。
第一种办法：
<% 
Function IsValidEmail(Email) 
ValidFlag = False 
If (Email <> "") And (InStr(1, Email, "@") > 0) And (InStr(1, Email, ".") > 0) Then 
atCount = 0 
SpecialFlag = False 
For atLoop = 1 To Len(Email) 
atChr = Mid(Email, atLoop, 1) 
If atChr = "@" Then atCount = atCount + 1 
If (atChr >= Chr(32)) And (atChr <= Chr(44)) Then SpecialFlag = True 
If (atChr = Chr(47)) Or (atChr = Chr(96)) Or (atChr >= Chr(123)) Then SpecialFlag = True 
If (atChr >= Chr(58)) And (atChr <= Chr(63)) Then SpecialFlag = True 
If (atChr >= Chr(91)) And (atChr <= Chr(94)) Then SpecialFlag = True 
Next 
If (atCount = 1) And (SpecialFlag = False) Then 
BadFlag = False 
tAry1 = Split(Email, "@") 
UserName = tAry1(0) 
DomainName = tAry1(1) 
If (UserName = "") Or (DomainName = "") Then BadFlag = True 
If Mid(DomainName, 1, 1) = "." then BadFlag = True 
If Mid(DomainName, Len(DomainName), 1) = "." then BadFlag = True 
ValidFlag = True 
' 格式正确返回Ture
End If 
End If 
If BadFlag = True Then ValidFlag = False 
' 格式不正确返回False
IsValidEmail = ValidFlag 
End Function 
%> 
    第二种办法：
<%
function IsValidEmail(email)
dim names, name, i, c
IsValidEmail = true
names = Split(email, "@")
if UBound(names) <> 1 then
  IsValidEmail = false
  exit function
end if
for each name in names
  if Len(name) <= 0 then
    IsValidEmail = false
    exit function
  end if
  for i = 1 to Len(name)
    c = Lcase(Mid(name, i, 1))
    if InStr("abcdefghijklmnopqrstuvwxyz_-.", c) <= 0 and not 
IsNumeric(c) then
      IsValidEmail = false
      exit function
    end if
  next
  if Left(name, 1) = "." or Right(name, 1) = "." then
      IsValidEmail = false
      exit function
  end if
next
if InStr(names(1), ".") <= 0 then
  IsValidEmail = false
  exit function
end if
i = Len(names(1)) - InStrRev(names(1), ".")
if i <> 2 and i <> 3 then
  IsValidEmail = false
  exit function
end if
if InStr(email, "..") > 0 then
  IsValidEmail = false
end if
end function
%>
    第三种办法，用下面这个函数进行判断。它会检查邮件地址是否含有“@”，以及“.”是否在“@”后面：
function isEmail(pInString)
  lAt = False
  lDot = false
  
  for x = 2 to len(pInstring)-1
    if mid(pInString,x,1) = "@" then lAt = True
      if mid(pInString,x,1) = "." and lAt = True then lDot = True
  next
  
  if lAt = True and lDot = True then 
    isEmail = True
  else
    isEmail = False
  end if
end function

上一个：如何判断用户在某一页面逗留了多长时间？
下一个：如何利用数据绑定来处理XML文件？