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如何判断电子邮件的地址格式是否正确？

时间：2022-03-22 02:38 编辑：来源：阅读：
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摘要：如何判断电子邮件的地址格式是否正确？

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

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