如何截取String,Eamil中的用户名 比如,字符串,[email protected]想截取ABC应该用什么函数.请教了. 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 不太清楚,好像这样写"[email protected]".substring(0,"[email protected]".indexOf("@")); 试试这个/* * email 取用户名 * @author sundful */public class EmailTest { public static void main(String[] args) { String s="[email protected]"; getUser(s); } public static void getUser(String s) { if(s==null || s.trim().length()<0) return; int num=s.indexOf("@"); if(num>0) { String ss=s.substring(0,num); System.out.println(ss); } }} package test;public class TT { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub String str = "[email protected]"; int index = str.indexOf("@"); String result = str.substring(0, index); System.out.println(result); }} package demo;public class CutString { public static void main(String[] args){ new CutString().doSplit(); } public void doSplit() { String email = "[email protected]"; String[] splits = email.split("@"); System.out.println(splits[0]); }} public static void main(String[] args){ String str1="[email protected]"; String[] sz=str1.split("@"); System.out.println(sz[0]); String s="[email protected]"; } String str="[email protected]";String[] result=str.split("@");System.out.println(result[0]); String stra = [email protected];String[] strarr = stra.split("@");System.out.println(strarr[0].toString()); Thank you so much!!!!!! 建议用正则表达式。 import java.util.regex.Pattern;import java.util.regex.Matcher;public class Test {public static void main(String[] args) {if(args.length != 1) {System.out.println("请输入一个参数!");System.exit(0);}Pattern p = Pattern.compile("\\w @(\\w .) [a-z]{2,3}");Matcher m = p.matcher(args[0]);boolean b = m.matches();if(b) {System.out.println("有效邮件地址");} else {System.out.println("无效邮件地址");}}}Pls see up codes. 额......貌似11楼表达式有问题....假设..我的EMAIL地址为:[email protected]就显示为错了.....应该是("\\w @(\\w .) [a-z]+")吧 没毛病邮箱好像没有4位结尾的常见的点.cn .com .name是什么 14楼说的也对,确实有.info .name之类的域名,不过在本题中检测邮箱地址是否正确用我的那个是没问题的。 另:用indexOf()和split()都是很好的方法! 没时间了 写个简单点的,主要是用group(1)import java.util.regex.Matcher;import java.util.regex.Pattern;public class GetEmailName { public static void main(String[] args) { String str = "[email protected]"; String result = ""; Pattern pattern = Pattern.compile("(\\w*)@\\w*\\.\\w*"); Matcher matcher = pattern.matcher(str); if(matcher.find()) System.out.println(matcher.group(1)); }} public class Test{ public static void main(String[] args){ String s="[email protected]"; StringBuffer br=new StringBuffer(); for(int i=0;i<=s.length();i++){ if(String.valueOf(s.charAt(i)).equals("@")){ break; } br.append(s.charAt(i)); } System.out.println(br); }} 再贴一个好点的这里用到是正则表达式的捕获组import java.util.regex.Matcher;import java.util.regex.Pattern;public class GetEmailName { public static void main(String[] args) { String str = "[email protected]"; String result = ""; Pattern pattern = Pattern.compile("^([a-z0-9]+(.[_a-z0-9-]+)*)@[a-z0-9-]+(.[a-z0-9-]+)*$"); Matcher matcher = pattern.matcher(str); if(matcher.find()) System.out.println(matcher.group(1)); }} int index=ss.indexof("@");String s=ss.subString(0,index); public static void main(String[] args) { // TODO Auto-generated method stub String str = "[email protected]"; Pattern pa = Pattern.compile(".*(?=@)"); Matcher matcher = pa.matcher(str); while (matcher.find()) { System.out.println(matcher.group()); } } 求高手帮助看看下面代码 求线程高人指点 数字的格式问题 java虚拟机将方法体中被解释器转换后的一条条指令存放在什么地方啊 请教:我在resin下 用jspSmartUpload.jar 进行上传,总是报错?在tomcat下上传都是成功的 编写jsp文件,帮忙推荐几个比较好的工具 jstl中$的问题?谢谢! 线程休眠 java两个接口,合写成一个???? Scanner读取文件问题 当对象比较大引起OutOfMemory错误时该怎么处理呢 将要去华为面试。
* email 取用户名
* @author sundful
*/
public class EmailTest {
public static void main(String[] args) {
String s="[email protected]";
getUser(s);
} public static void getUser(String s) {
if(s==null || s.trim().length()<0) return;
int num=s.indexOf("@");
if(num>0)
{
String ss=s.substring(0,num);
System.out.println(ss);
}
}
}
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub String str = "[email protected]";
int index = str.indexOf("@");
String result = str.substring(0, index);
System.out.println(result); }
}
new CutString().doSplit();
}
public void doSplit() {
String email = "[email protected]";
String[] splits = email.split("@");
System.out.println(splits[0]);
}
}
public static void main(String[] args){ String str1="[email protected]";
String[] sz=str1.split("@");
System.out.println(sz[0]);
String s="[email protected]";
}
String[] result=str.split("@");
System.out.println(result[0]);
String[] strarr = stra.split("@");
System.out.println(strarr[0].toString());
import java.util.regex.Pattern;
import java.util.regex.Matcher;public class Test {
public static void main(String[] args) {if(args.length != 1) {
System.out.println("请输入一个参数!");
System.exit(0);
}
Pattern p = Pattern.compile("\\w @(\\w .) [a-z]{2,3}");
Matcher m = p.matcher(args[0]);
boolean b = m.matches();
if(b) {
System.out.println("有效邮件地址");
} else {
System.out.println("无效邮件地址");
}
}
}Pls see up codes.
假设..我的EMAIL地址为:
[email protected]
就显示为错了.....应该是
("\\w @(\\w .) [a-z]+")
吧
import java.util.regex.Pattern;public class GetEmailName {
public static void main(String[] args) {
String str = "[email protected]";
String result = "";
Pattern pattern = Pattern.compile("(\\w*)@\\w*\\.\\w*");
Matcher matcher = pattern.matcher(str);
if(matcher.find())
System.out.println(matcher.group(1));
}}
public class Test{
public static void main(String[] args){
String s="[email protected]";
StringBuffer br=new StringBuffer();
for(int i=0;i<=s.length();i++){
if(String.valueOf(s.charAt(i)).equals("@")){
break;
}
br.append(s.charAt(i));
}
System.out.println(br);
}
}
这里用到是正则表达式的捕获组import java.util.regex.Matcher;
import java.util.regex.Pattern;public class GetEmailName { public static void main(String[] args) {
String str = "[email protected]";
String result = "";
Pattern pattern = Pattern.compile("^([a-z0-9]+(.[_a-z0-9-]+)*)@[a-z0-9-]+(.[a-z0-9-]+)*$");
Matcher matcher = pattern.matcher(str);
if(matcher.find())
System.out.println(matcher.group(1));
}}
String s=ss.subString(0,index);
// TODO Auto-generated method stub
String str = "[email protected]";
Pattern pa = Pattern.compile(".*(?=@)");
Matcher matcher = pa.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}
}