如何判斷電子郵件的地址格式是否正確?
第一種辦法:
<%
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
' 格式正確返回Ture。
End If
End If
If BadFlag = True Then ValidFlag = False
' 格式不正確返回False。
IsValidEmail = ValidFlag
End Function
%>
<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />
第二種辦法:
<%
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
%>
第三種辦法,用下面這個(gè)函數(shù)進(jìn)行判斷。它會(huì)檢查郵件地址是否含有“@”,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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