如何判断电子邮件的地址格式是否正确？

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 = 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