[String]提取字符串子串的方法 String s = "abcd@1234";String ss[] = s.split("@"); 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 顺手写的,不一定对String s = new String("abcd@1234");int pos = s.indexOf("@");String s1 = s.substring(0,pos-1);String s2 = s.substring(pos,s.length()); 同意 disller的方法: String s = "abcd@1234"; String ss[] = s.split("@"); System.out.println(ss[0]); //打印@前面字符串 System.out.println(ss[1]); //打印@后面字符串 String s = new String("abcd@1234");int pos = s.indexOf("@");String s1 = s.substring(0,pos);String s2 = s.substring(pos + 1,s.length()); StringTokenizer这个类可以解决你的问题,StringTokenizer paramsTokenizer = new StringTokenizer(yourinput,"@"); int size = paramsTokenizer.countTokens(); for(int i=0;i<size;i++) { String helper = paramsTokenizer.nextToken().toString(); }可依次取的@前后的子串 /** *@function 将指定间隔符的字符串拆分为数组 * *@param oriString String 需被拆分的字符串 * *@param separator String 间隔符 * *@return String[] 字符串数组 * *@usage: * String str = "a,b,c,d,e"; * final String separator = ","; * String[] result = toArray(str, separator); * */ public static String[] toArray(String oriString, String separator){ if(oriString == null) throw new NullPointerException("The parameter [string] cannot be null."); if(separator == null) throw new NullPointerException("The parameter [separator] cannot be null."); if(separator.length() == 0) throw new IllegalArgumentException("The parameter [separator] cannot be empty."); java.util.StringTokenizer st = new java.util.StringTokenizer(oriString, separator); java.util.List list = new java.util.LinkedList(); while(st.hasMoreTokens()){ list.add(st.nextToken()); } return (String[])list.toArray(new String[list.size()]); } String b = "1,2,3,4,5,6"; public static void main(String args[]){ Split split = new Split(); split.M2(); split.M1(); } protected void M2(){ String c[] = b.split(","); for(int i=0 ;i<c.length;i++){ System.out.println(c[i]); } } protected void M1(){ StringTokenizer paramsTokenizer = new StringTokenizer(b,","); int size = paramsTokenizer.countTokens(); for(int i=0;i<size;i++){ String helper = paramsTokenizer.nextToken().toString(); System.out.println(helper); } } JAVA 请问.的正则表达式怎样写? 问一个java小程序 一个选择题:java(J2EE)和管理员(AIX,Orical) 从键盘中如何读取输入数据?(最好能多个方法) 谁能给我一个executeBath()的例子吗 JAVA小菜鸟高分求一JAVA小程序 如果一个对象正在运行,但该对象不是线程,有什么方法结束这个对象的运行 10(int)+0.8(float)+3.0(double)=??? 请问一下那个不丢失数据,可以给硬盘分区的软件是什么? 我的applet运行时出现奇怪的问题(内详),请大家帮忙 急!String对象的问题,谢谢
int pos = s.indexOf("@");
String s1 = s.substring(0,pos-1);
String s2 = s.substring(pos,s.length());
String s = "abcd@1234";
String ss[] = s.split("@");
System.out.println(ss[0]); //打印@前面字符串
System.out.println(ss[1]); //打印@后面字符串
int pos = s.indexOf("@");
String s1 = s.substring(0,pos);
String s2 = s.substring(pos + 1,s.length());
int size = paramsTokenizer.countTokens();
for(int i=0;i<size;i++)
{
String helper = paramsTokenizer.nextToken().toString();
}
可依次取的@前后的子串
*@function 将指定间隔符的字符串拆分为数组
*
*@param oriString String 需被拆分的字符串
*
*@param separator String 间隔符
*
*@return String[] 字符串数组
*
*@usage:
* String str = "a,b,c,d,e";
* final String separator = ",";
* String[] result = toArray(str, separator);
*
*/
public static String[] toArray(String oriString, String separator){ if(oriString == null)
throw new NullPointerException("The parameter [string] cannot be null.");
if(separator == null)
throw new NullPointerException("The parameter [separator] cannot be null.");
if(separator.length() == 0)
throw new IllegalArgumentException("The parameter [separator] cannot be empty."); java.util.StringTokenizer st = new java.util.StringTokenizer(oriString, separator); java.util.List list = new java.util.LinkedList();
while(st.hasMoreTokens()){
list.add(st.nextToken());
} return (String[])list.toArray(new String[list.size()]);
}
public static void main(String args[]){
Split split = new Split();
split.M2();
split.M1();
}
protected void M2(){
String c[] = b.split(",");
for(int i=0 ;i<c.length;i++){
System.out.println(c[i]);
}
}
protected void M1(){
StringTokenizer paramsTokenizer = new StringTokenizer(b,",");
int size = paramsTokenizer.countTokens();
for(int i=0;i<size;i++){
String helper = paramsTokenizer.nextToken().toString();
System.out.println(helper);
}
}