如何判断电子邮件的地址格式是否正确？
我在网站上设置了邮件列表功能，实现用户自动订阅和发送，但很多用户输入和提交的邮件地址格式都是无效的，无法处理。请问如何解决这一问题？
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 = Fal
If (Email <> "") And (InStr(1, Email, "@") > 0) And (InStr(1, Email, ".") > 0) Then
atCount = 0
SpecialFlag = Fal
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 = Fal) Then
BadFlag = Fal
tAry1 = Split(Email, "@")
UrName = tAry1(0)
DomainName = tAry1(1)
If (UrName = "") 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 = Fal
' 格式不正确返回Fal
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 = fal
exit function
end if
for each name in names
if Len(name) <= 0 then
IsValidEmail = fal
exit function
end if
for i = 1 to Len(name)
c = Lca(Mid(name, i, 1))
if InStr("abcdefghijklmnopqrstuvwxyz_-.", c) <= 0 and not
IsNumeric(c) then
IsValidEmail = fal
exit function
end if
next
if Left(name, 1) = "." or Right(name, 1) = "." then
IsValidEmail = fal
exit function
end if
next
if InStr(names(1), ".") <= 0 then
IsValidEmail = fal
exit function
end if
i = Len(names(1)) - InStrRev(names(1), ".")
if i <> 2 and i <> 3 then
IsValidEmail = fal
exit function
end if
if InStr(email, "..") > 0 then
IsValidEmail = fal
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 = Fal
lDot = fal
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
el
isEmail = Fal
end if
end function
