Java 如何计算字符串中所有数字的总和

How to calculate sum of all numbers in a string

提问by user3337429

How do I calculate the sum of all the numbers in a string? In the example below, the expected result would be 4+8+9+6+3+5. My attempt is below. Also could I calculate the sum of only those numbers which are divisible by 2?

如何计算字符串中所有数字的总和？在下面的例子中，预期的结果是4+8+9+6+3+5。我的尝试如下。我也可以只计算那些可以被 2 整除的数字的总和吗？

int sum=0;
String s = "jklmn489pjro635ops";
for(int i=0; i<s.length(); i++) {
    char temp = s.charAt(i);
    if (Character.isDigit(temp)) {
        int b = Integer.parseInt(String.valueOf(temp));
        sum=sum+b;
    }
}
System.out.println(sum);

回答by VinayVeluri

Some thing like this..

像这样的东西..

import org.apache.commons.lang.StringUtils;

public static void main(String[] args) {

    String a = "jklmn489pjro635ops";

    int length = a.length();

    char c;
    int count = 0;
    for(int i=0;i<length;i++) {
        c  = a.charAt(i);
        if(StringUtils.isNumeric(String.valueOf(c))) {
            count = count + Integer.parseInt(String.valueOf(c));

        }
    }
    System.out.println(count);
}

回答by user2795559

int sum = 0;
for(int i = 0; i < str.length()-1;){
  try{
    sum += Integer.parseInt(str.substring(i, ++i));
  }catch(Exception e){}
}

回答by user2795559

    String input = "abc12xx3";
    Pattern pattern = Pattern.compile("[\d]");
    Matcher matcher = pattern.matcher(input);
    int sum = 0;
    while(matcher.find()) {
        sum += Integer.parseInt(matcher.group());
    }
    return sum;

回答by renke

Parsing chars back to Stringand then to Integeris too expensive, since you already have a char. You should try doing this:

将字符解析回 toString和 toInteger太昂贵了，因为您已经有了一个字符。你应该尝试这样做：

 String a = "jklmn489pjro635ops";
 int sum = 0;
 int evenSum = 0;
 for (char c : a.replaceAll("\D", "").toCharArray()) {
     int digit = c - '0';
     sum += digit;
     if (digit % 2 == 0) {
         evenSum += digit;
     }
 }
 System.out.println(sum);
 System.out.println(evenSum);

回答by Jess

Here's a full mainline example you can run. This should answer your question and all the follow on questions in the comments.

这是您可以运行的完整主线示例。这应该回答您的问题以及评论中的所有问题。

import java.util.regex.*;
public class StringSum {
    public static void main(String[] args) {
        int sum = 0;
        Pattern pattern = Pattern.compile("([2468])"); // Who do we appreciate!
        Matcher matcher = pattern.matcher(args[0]);
        while (matcher.find()) {
            sum += Integer.parseInt(matcher.group(1));
        }
        System.out.println(sum);
    }
}

Some example executions:

一些示例执行：

$ javac StringSum.java && java StringSum jklmn489pjro635ops
18
$ javac StringSum.java && java StringSum a1b2c3d4
6

回答by Jess

int banana = 0;
String meow = "74387493";

for (int i = 0; i < meow.length(); i++) 
{
    String cows = ""+meow.charAt(i);
    int b =  Integer.parseInt(cows);
    banana = banana + b;
}

System.out.println(banana);

The above code will add all the numbers in a string (sorry for the silly names).

上面的代码将添加一个字符串中的所有数字（对不起，愚蠢的名字）。

回答by vbh

String a = "jklmn489pjro635ops";

int sum = 0;

for(int i = 0; i < a.length(); i++) {
    if(Character.isDigit(a.charAt(i))) {
        sum = sum + Integer.parseInt(a.charAt(i) + "");
    } 
}

System.out.println(sum);

回答by Dark Army

Scanner sc = new Scanner(System.in);

    System.out.println("Enter the String");

    String input = sc.next();

    int sum = 0;

    char[] c = input.toCharArray();

    for(int i = 0;i<=c.length-1;i++)
    {
      if(Character.isDigit(c[i]))
      {
          Character c1 = c[i];
          String s1 = c1.toString();
          int i1 = Integer.parseInt(s1);

          sum = sum+i1;
      }
    }
    System.out.println(sum);

回答by Kundan Shinde

This could be a simple and efficient way:

这可能是一种简单而有效的方法：

static int sumFromString(String s){
       int sum = 0;
        for(int i = 0; i < s.length() ; i++){
            if( Character.isDigit(s.charAt(i)) ){
                sum = sum + Character.getNumericValue(s.charAt(i));
            }
        }
        return sum;
    }

Another way was to convert char to String and then to int, which is not so efficient:

另一种方法是将 char 转换为 String，然后转换为 int，这样效率不高：

sum = sum + Integer.parseInt(s.charAt(i) + "");

回答by Asim Maqsood

I think this would be the simplest and easiest way

我认为这将是最简单和最简单的方法

static int sumOfString(String str){
    int sum = 0;
    for(int i=0; i<str.length(); i++){              
            if(Character.isDigit(str.charAt(i))){
                    sum+=str.charAt(i)-'0';
                }
        }
    return sum;
}