我想通过分隔符“-”将字符串“004-034556”拆分为两个字符串:

part1 = "004";
part2 = "034556";

这意味着第一个字符串将包含“-”之前的字符,第二个字符串将包括“-”之后的字符。

我还想检查字符串中是否有“-”。


当前回答

String str="004-034556"
String[] sTemp=str.split("-");// '-' is a delimiter

string1=004 // sTemp[0];
string2=034556//sTemp[1];

其他回答

直接处理字符串的另一种方法是将正则表达式与捕获组一起使用。这样做的优点是,它可以直接暗示对输入的更复杂的约束。例如,以下命令将字符串拆分为两部分,并确保两者仅由数字组成:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

class SplitExample
{
    private static Pattern twopart = Pattern.compile("(\\d+)-(\\d+)");

    public static void checkString(String s)
    {
        Matcher m = twopart.matcher(s);
        if (m.matches()) {
            System.out.println(s + " matches; first part is " + m.group(1) +
                               ", second part is " + m.group(2) + ".");
        } else {
            System.out.println(s + " does not match.");
        }
    }

    public static void main(String[] args) {
        checkString("123-4567");
        checkString("foo-bar");
        checkString("123-");
        checkString("-4567");
        checkString("123-4567-890");
    }
}

由于模式在本例中是固定的,因此可以预先编译并存储为静态成员(在示例中是在类加载时初始化的)。正则表达式为:

(\d+)-(\d+)

括号表示捕获组;可以通过Match.group()方法访问与正则表达式的该部分匹配的字符串,如图所示。\d匹配一个十进制数字,+表示“匹配一个或多个前一个表达式)。-没有特殊含义,因此只匹配输入中的字符。请注意,当将其写成Java字符串时,需要对反斜杠进行双转义。其他一些示例:

([A-Z]+)-([A-Z]+)          // Each part consists of only capital letters 
([^-]+)-([^-]+)            // Each part consists of characters other than -
([A-Z]{2})-(\d+)           // The first part is exactly two capital letters,
                           // the second consists of digits

可以使用Split():

import java.io.*;

public class Splitting
{

    public static void main(String args[])
    {
        String Str = new String("004-034556");
        String[] SplittoArray = Str.split("-");
        String string1 = SplittoArray[0];
        String string2 = SplittoArray[1];
    }
}

否则,您可以使用StringTokenizer:

import java.util.*;
public class Splitting
{
    public static void main(String[] args)
    {
        StringTokenizer Str = new StringTokenizer("004-034556");
        String string1 = Str.nextToken("-");
        String string2 = Str.nextToken("-");
    }
}

使用模式是Java 8。下面是路!!

package com.company;

import java.util.regex.Pattern;

public class umeshtest {

    public static void main(String a[]) {
        String ss = "I'm Testing and testing the new feature";
        Pattern.compile(" ").splitAsStream(ss).forEach(s -> System.out.println(s));
    }
}

消耗最少资源的最快方式可能是:

String s = "abc-def";
int p = s.indexOf('-');
if (p >= 0) {
    String left = s.substring(0, p);
    String right = s.substring(p + 1);
} else {
  // s does not contain '-'
}
// This leaves the regexes issue out of question
// But we must remember that each character in the Delimiter String is treated
// like a single delimiter        

public static String[] SplitUsingTokenizer(String subject, String delimiters) {
   StringTokenizer strTkn = new StringTokenizer(subject, delimiters);
   ArrayList<String> arrLis = new ArrayList<String>(subject.length());

   while(strTkn.hasMoreTokens())
      arrLis.add(strTkn.nextToken());

   return arrLis.toArray(new String[0]);
}