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如何判断电子邮件的地址格式是否正确？

2017年12月08日  | 移动技术网IT编程  | 我要评论

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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