Question

我想在字符串中自动增加整数（是产品代码）例如：ABC00001 - ＆gt; ABC00002; ABC00009 - ＆gt; ABC00010 ....和ABC99999 - ＆gt; ABC000001 ... 它只是增加整数。我不知道如何解决它，因为它有数字＆＃34; 0000＆＃34;。我搜索一个函数：

public static void main(String[] args) {

   Pattern digitPattern = Pattern.compile("(\\d)"); // EDIT: Increment each digit.

   Matcher matcher = digitPattern.matcher("test001check2");
   StringBuffer result = new StringBuffer();
   while (matcher.find())
   {
       matcher.appendReplacement(result, String.valueOf(Integer.parseInt(matcher.group(1)) + 1));
   }
   matcher.appendTail(result);
   System.out.println(result.toString());

  }

但它会使每个数字都折皱。我该如何解决？

Answer 1

如果我理解你的问题，一种可能的方法是从String获取前3个字符，然后将其他所有字符提取到int。然后，您可以使用类似

之类的内容格式化输出（并将其提取到方法中）

private static String increment(String str) {
    String base = str.substring(0, 3);
    int val = Integer.parseInt(str.substring(4));
    return String.format("%s%06d", base, val + 1);
}

然后测试它

public static void main(String[] args) {
    String str = "ABC000001";
    System.out.println(str);
    System.out.println(increment(str));
}

输出

ABC000001
ABC000002

Answer 2

public static void main(String[] args) {
   // use \d+ to match the whole number
   Pattern digitPattern = Pattern.compile("(\\d+)");

   Matcher matcher = digitPattern.matcher("test001check2");
   StringBuffer result = new StringBuffer();
   while (matcher.find())
   {
      String match = matcher.group(1);
      int numDigits = match.length(); // get number of digits in the string
      int newValue = Integer.parseInt(matcher.group(1)) + 1; // get the raw value of the string, add one
      // pad the new value with the right number of zeros, so 001 will become 002 and not just 2
      String newValueStr = String.format("%0" + numDigits + "d", newValue);
      matcher.appendReplacement(result, newValueStr);
   }
   matcher.appendTail(result);
   System.out.println(result.toString());

  }

我机器上的输出：test002check3

Answer 3

好的，抱歉我的问题不清楚。我只想增加mix字符串中的最后一个整数。例如：test001check2 - ＆gt; test001check3; test001check9-＆GT; test001check00; test001check09 - ＆gt; test001check10; test001check99 - ＆gt; test001check000 .... 现在我发现了一个具有上述要求的功能。

public final char MIN_DIGIT = '0';
public final char MAX_DIGIT = '9';

public String incrementedCode(String original) {
    if(original != null && !"".equals(original.trim())){
        StringBuilder buf = new StringBuilder(original);
        int i = buf.length() -1;
        while(i >= 0) {
           char c = buf.charAt(i);
           if(c >= MIN_DIGIT && c<= MAX_DIGIT){
               c++;
               if(c > MAX_DIGIT) { // overflow, carry one
                  if(buf.charAt(i - 1) > MAX_DIGIT){
                      buf.setCharAt(i, MIN_DIGIT);
                      buf.insert(i, "0");
                      return buf.toString();
                  }
                  buf.setCharAt(i, MIN_DIGIT);
                  i--;
                  continue;
               }
               buf.setCharAt(i, c);
               return buf.toString();
           }
           i--;
        }
        // overflow at the first "digit", need to add one more digit
        buf.insert(0, MIN_DIGIT);
        return buf.toString();
    }else{
        return "";
    }
}

mix string中的增量整数包括：string + integer

3 个答案: