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);

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);

如何(^ \ \ d) * ([0 - 9] + [\ \ s] * [,] {0,1} \ \ [s] *[0 - 9] *)。*我认为它可以处理小数部分的数字。 我加入了空白和可能的分隔符。 我试图得到的数字从一个字符串，包括浮动，并考虑到用户可能会犯一个错误，并包括空白，而键入的数字。

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);
}

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());
         }

Pattern p = Pattern.compile("(\\D+)(\\d+)(.*)");
Matcher m = p.matcher("this is your number:1234 thank you");
if (m.find()) {
    String someNumberStr = m.group(2);
    int someNumberInt = Integer.parseInt(someNumberStr);
}