如何判断一个String是否符合e_mail格式??? 如题 解决方案 » 免费领取超大流量手机卡,每月29元包185G流量+100分钟通话, 中国电信官方发货 最简单的办法就是String a;if(a.indexof("@")>0) return true;else return false;要更加严谨的就用正则表达式 package Csdn;public class HelloWorld { boolean isEmail (String e){ if(e.indexOf("@")==-1||e.indexOf("@")==e.length()-1||e.indexOf("@")==0||e.lastIndexOf("@")!=e.indexOf("@")) return false; else return true; } public static void main(String []args){ String email ="112jjkj";//@不能第一,最后,重复,不出现 HelloWorld h = new HelloWorld(); System.out.println(h.isEmail(email)); }} String mail="[email protected]"; boolean d = mail.matches("^([a-z0-9A-Z]+[-|\\.]?)+[a-z0-9A-Z]@([a-z0-9A-Z]+(-[a-z0-9A-Z]+)?\\.)+[a-zA-Z]{2,}$"); System.out.println(d); /** * 验证Email格式是否正确 * @param email 一个email字符串参数 * @return 如果email格式正确则返回true,否则返回false */ public boolean checkEmail(String email){ boolean flag=false; Pattern p=Pattern.compile("^[a-zA-Z0-9_]+@\\w+\\.[a-zA-Z]+(\\.[a-zA-Z]+)*$"); Matcher m = null; m = p.matcher(email); flag = m.matches(); return flag; } > lip009(深蓝忧郁) ( ) 信誉:100 Blog 的方法最好了! lip009(深蓝忧郁)的方法有漏洞吧:[email protected]也是合法的啊,呵呵 楼上的有没有测试过哦...我就很崩溃[email protected]这个能通过才鬼了 告诉你个正则表达式的好网站:http://www.regular-expressions.info/examples.html^[A-Z0-9._%-]+@[A-Z0-9.-]+\.(?:[A-Z]{2}|com|org|net|biz|info|name|aero|biz|info|jobs|museum|name)$ 这个符合RFC 822 的:(?:(?:\r\n)?[ \t])*(?:(?:(?:[^()<>@,;:\\".\[\] \x00-\x1F]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[ \t]))*"(?:(?:\r\n)?[ \t])*)(?:\.(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \x00-\x1F]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[ \t]))*"(?:(?:\r\n)?[ \t])*))*@(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \x00-\x1F]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|\[([^\[\]\r\\]|\\.)*\](?:(?:\r\n)?[ \t])*)(?:\.(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \x00-\x1F]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|\[([^\[\]\r\\]|\\.)*\](?:(?:\r\n)?[ \t])*))*|(?:[^()<>@,;:\\".\[\] \x00-\x1F]+(?:(?:(?:\r\n)?[ 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String a;
if(a.indexof("@")>0)
return true;
else
return false;要更加严谨的就用正则表达式
boolean isEmail (String e){
if(e.indexOf("@")==-1||e.indexOf("@")==e.length()-1||e.indexOf("@")==0||e.lastIndexOf("@")!=e.indexOf("@"))
return false;
else
return true;
}
public static void main(String []args){
String email ="112jjkj";//@不能第一,最后,重复,不出现
HelloWorld h = new HelloWorld();
System.out.println(h.isEmail(email));
}}
boolean d = mail.matches("^([a-z0-9A-Z]+[-|\\.]?)+[a-z0-9A-Z]@([a-z0-9A-Z]+(-[a-z0-9A-Z]+)?\\.)+[a-zA-Z]{2,}$");
System.out.println(d);
* 验证Email格式是否正确
* @param email 一个email字符串参数
* @return 如果email格式正确则返回true,否则返回false
*/
public boolean checkEmail(String email){
boolean flag=false;
Pattern p=Pattern.compile("^[a-zA-Z0-9_]+@\\w+\\.[a-zA-Z]+(\\.[a-zA-Z]+)*$");
Matcher m = null;
m = p.matcher(email);
flag = m.matches(); return flag;
}
[email protected]也是合法的啊,呵呵
我就很崩溃[email protected]这个能通过才鬼了
http://www.regular-expressions.info/examples.html
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