试着这样做:

Pattern p = Pattern.compile("^.+(\\d+).+");
Matcher m = p.matcher("Testing123Testing");

if (m.find()) {
    System.out.println(m.group(1));
}

简单的解决方案

// Regexplanation:
// ^       beginning of line
// \\D+    1+ non-digit characters
// (\\d+)  1+ digit characters in a capture group
// .*      0+ any character
String regexStr = "^\\D+(\\d+).*";

// Compile the regex String into a Pattern
Pattern p = Pattern.compile(regexStr);

// Create a matcher with the input String
Matcher m = p.matcher(inputStr);

// If we find a match
if (m.find()) {
    // Get the String from the first capture group
    String someDigits = m.group(1);
    // ...do something with someDigits
}

一个Util类的解

public class MyUtil {
    private static Pattern pattern = Pattern.compile("^\\D+(\\d+).*");
    private static Matcher matcher = pattern.matcher("");

    // Assumptions: inputStr is a non-null String
    public static String extractFirstNumber(String inputStr){
        // Reset the matcher with a new input String
        matcher.reset(inputStr);

        // Check if there's a match
        if(matcher.find()){
            // Return the number (in the first capture group)
            return matcher.group(1);
        }else{
            // Return some default value, if there is no match
            return null;
        }
    }
}

...

// Use the util function and print out the result
String firstNum = MyUtil.extractFirstNumber("Testing4234Things");
System.out.println(firstNum);

在Java 1.4及以上版本中:

String input = "...";
Matcher matcher = Pattern.compile("[^0-9]+([0-9]+)[^0-9]+").matcher(input);
if (matcher.find()) {
    String someNumberStr = matcher.group(1);
    // if you need this to be an int:
    int someNumberInt = Integer.parseInt(someNumberStr);
}

完整的例子:

private static final Pattern p = Pattern.compile("^([a-zA-Z]+)([0-9]+)(.*)");
public static void main(String[] args) {
    // create matcher for pattern p and given string
    Matcher m = p.matcher("Testing123Testing");

    // if an occurrence if a pattern was found in a given string...
    if (m.find()) {
        // ...then you can use group() methods.
        System.out.println(m.group(0)); // whole matched expression
        System.out.println(m.group(1)); // first expression from round brackets (Testing)
        System.out.println(m.group(2)); // second one (123)
        System.out.println(m.group(3)); // third one (Testing)
    }
}

由于您正在寻找第一个数字，您可以使用这样的regexp:

^\D+(\d+).*

m.group(1)将返回第一个数字。注意，带符号的数字可以包含负号:

^\D+(-?\d+).*

如果你正在从文件中读取，那么这可以帮助你

              try{
             InputStream inputStream = (InputStream) mnpMainBean.getUploadedBulk().getInputStream();
             BufferedReader br = new BufferedReader(new InputStreamReader(inputStream));
             String line;
             //Ref:03
             while ((line = br.readLine()) != null) {
                if (line.matches("[A-Z],\\d,(\\d*,){2}(\\s*\\d*\\|\\d*:)+")) {
                     String[] splitRecord = line.split(",");
                     //do something
                 }
                 else{
                     br.close();
                     //error
                     return;
                 }
             }
                br.close();

             }
         }
         catch (IOException  ioExpception){
             logger.logDebug("Exception " + ioExpception.getStackTrace());
         }

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

public class Regex1 {
    public static void main(String[]args) {
        Pattern p = Pattern.compile("\\d+");
        Matcher m = p.matcher("hello1234goodboy789very2345");
        while(m.find()) {
            System.out.println(m.group());
        }
    }
}

输出:

1234
789
2345