如何将电子邮件地址子串

时间：2014-09-09 20:14:39

标签： c# string email substring

我是C＃的新手，不知道如何在电子邮件地址中获取子字符串。

例如，如果电子邮件地址是：name.surname@company.com或support@company.com 我想知道如何将此信息与此电子邮件地址分开？

与name.surname@company.com一样

string _name = name;
string _surname = surname;
string _company = company;
string domaine = company.com;

或者如果它只是support@company.com那么

string _name = support;
string _company = company;
string domaine = company.com

感谢您的帮助

3 个答案:

答案 0 :(得分：5)

您可以使用MailAddress课程。

var mail = new MailAddress("name.surname@company.com");
var user = mail.User; // name.surname
var host = mail.Host; // company.com

答案 1 :(得分：2)

又一个选择：

string email = "thisemail@hotmail.com";

string user = email.Substring(0, email.IndexOf("@"));
string domain = email.Substring(email.IndexOf("@")+1);
string company = string.Empty;

if (domain.Contains(".")) //just in case no '.'
{
    company = domain.Substring(0, domain.IndexOf("."));
}
else
{
    company = domain;
}

答案 2 :(得分：0)

解析电子邮件地址......比您想象的要复杂一点。 My answer问题C# Email Address validation可能对您有用。它不会匹配所有合法的电子邮件地址，但它应该匹配您通常会看到的每个地址。​​

如果您使用我在上面的答案中构建的正则表达式，您应该能够像这样提取local-part：

Match  m         = RxEmailAddress.Match( "foo@bar.com" ) ;
string localPart = m.Success ? m.Groups["localpart"].Value : null ;

这是获得你想要的另一种方式。我们编写一个正则表达式来将local-part分解成段。我们正在使用选项RegexOptions.ExplicitCapture来控制我们捕获的内容。在这种情况下，只有本地部分的单词。

Regex rx = new Regex( @"^(?<first>[^.@]+)(\.(?<next>[^.@]+))*@.+$" , RegexOptions.ExplicitCapture ) ;

有了这个，我们可以匹配一个电子邮件地址，因此：

Match m  = rx.Match( "foo.bar.bazbat.schazam@domain.com" ) ;

假设成功，我们可以将local-part的每个段解析成一个数组，如下所示：

if ( m.Success )
{
  string[] segments = m.Groups.Cast<Group>()
                      .Skip(1)
                      .SelectMany( g => g.Captures.Cast<Capture>() )
                      .Select( c => c.Value )
                      .ToArray()
                      ;
}

在上面的代码中，

  • 我们跳过第一组，因为它包含整场比赛，并且没那么有趣。
  • 我们使用SelectMany()收集每个匹配组的捕获列表，然后将生成的列表列表展平为一个捕获列表。
  • 然后我们选择每个捕获的值，并
  • 将其泵入阵列。

FWIW，这里是正则表达式，以匹配完整的RFC 822电子邮件地址。该标准自RFC 822起已经发展，因此可能不完整：

(?:(?:\r\n)?[ \t])*(?:(?:(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?:(?:\r\n)?[ \t]
)+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[ \t]))*"(?:(?:
\r\n)?[ \t])*)(?:\.(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?:(
?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[ 
\t]))*"(?:(?:\r\n)?[ \t])*))*@(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \000-\0
31]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|\[([^\[\]\r\\]|\\.)*\
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