Question

可以将字符串数组的所有值添加为正数或负数，然后将它们传递给双数组。

实施例：和价值 String sa_notas [] = {1,5，-2} 结果（1 + 5-2）= 4

错误消息：在org.apache.harmony.luni.util.FloatingPointParser.initialParse（FloatingPointParser.java:149）在org.apache.harmony.luni.util.FloatingPointParser.parseDouble（FloatingPointParser.java:281）在java.lang.Double.parseDouble（Double.java:318）在es.amedidaapp.gado.D_Anadir_Registro $ 3.afterTextChanged（D_Anadir_Registro.java:176）

感谢您的帮助

代码：

if (s.length() > 0) {
        s_contains_notas = et_notas.getText().toString();
        if (b_calculadora) {

            if (s_contains_notas.length() <= 0 || !s_contains_notas.matches("[^0-9]+")) {
                s_contains_notas = s_contains_notas.replace("+", " ");//separate positives numbers
                s_contains_notas = s_contains_notas.replaceAll("[^0-9, -]", ""); //delete letters
                s_contains_notas = s_contains_notas.replace(",", "."); //change format numbers
                s_contains_notas = s_contains_notas.replace("-", " -"); //separate negatives numbers

                String sa_notas[] = s_contains_notas.split(" "); //{1, 5, -2}
                double[] d_notas = new double[sa_notas.length];
                for (int i = 0; i < d_notas.length; i++) {
                    d_notas[i] = Double.parseDouble(sa_notas[i]); //ERROR to write negative numbers
                }
                double d_sum_notas = 0;
                for (Double d : d_notas) {
                    d_sum_notas += d;
                }
                s_euros = String.valueOf(d_sum_notas);
                et_euros.setText(nf_locale.format(d_sum_notas));
            }
        }
    }
}

Answer 1

一个不同的方法怎么样？

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


public class RegEx {

  // Example input:
  final static private String NUMBERS = "Data: 0,5€ -0.1$ 9 -3  #3";

  // Regular expression to match floating point numbers:
  final static private Pattern PATTERN = Pattern.compile("[-]?[0-9]+(?:[,.][0-9]+)?");

  public static void main(String[] args) {

    final Matcher m = PATTERN.matcher(NUMBERS);

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

  }

}

输出：

0,5
-0.1
9
-3
3

Answer 2

感谢JimmyB指导我，

下一个问题，新代码无法生成小数。

新代码

 if (s_contains_notas.length() <= 0 || !s_contains_notas.matches("[^0-9]+")) {

                List<String> list = new ArrayList<>();
                Pattern PATTERN = Pattern.compile("[-]?[0-9]+(?:[,.][0-9]+)?");
                Matcher m = PATTERN.matcher(s_contains_notas);
                while ( m.find() ) {
                    list.add(m.group());
                }

                double[] d_notas = new double[list.size()]; //create an array with the size of the list

                for (int i = 0; i < list.size(); ++i) { //iterate over the elements of the list
                    d_notas[i] = Double.parseDouble(list.get(i)); //store each element as a double in the array
                }

                double d_sum_notas = 0;
                for (Double d : d_notas) {
                    d_sum_notas += d;
                }
                s_euros = String.valueOf(d_sum_notas);
                et_euros.setText(nf_locale.format(d_sum_notas));

Answer 3

问题解决了

CODE：

 if (s_contains_notas.length() <= 0 || !s_contains_notas.matches("[^0-9]+")) {

                list = new ArrayList<>();
                Pattern PATTERN = Pattern.compile("[-]?[0-9]+(?:[,.][0-9]+)?");

                s_contains_notas = s_contains_notas.replaceAll("," , ".");
                Matcher m = PATTERN.matcher(s_contains_notas);
                while (m.find()) {
                    list.add(m.group());

                }

                double[] d_notas = new double[list.size()]; //create an array with the size of the failList

                for (int i = 0; i < list.size(); ++i) { //iterate over the elements of the list
                    d_notas[i] = Double.parseDouble(list.get(i)); //store each element as a double in the array
                }

                double d_sum_notas = 0;
                for (Double d : d_notas) {
                    d_sum_notas += d;
                }

                s_euros = String.valueOf(d_sum_notas);
                et_euros.setText(nf_locale.format(d_sum_notas));
            }

输出： 1 -1 0.5 -0.5 0,5

解析字符串加倍[]

3 个答案: