字符串转换为数字时出现NumberFormatException错误

用Integer.parseInt()转换字符时抛出NumberFormatException错误

原因：

一. 字符串中包含空格或非数字字符：检查所输入的字符串或者从字符串中提取数字

1 String类提供的方法：

public String stringToNumber (String string){
        string = string.trim();
        String string1 = "";
        if (string != null && !string.equals("")){
            for (int i = 0;i < string.length();i++){
                if (string.charAt(i) >= 48 && string.charAt(i) <= 57){
                    string1 += string.charAt(i);
                }
            }
        }
        return string1;
    }

2 正则表达式：

public String stringToNumber2(String string){
        String regular = "^[0-9]$";
        Pattern pattern = Pattern.compile(regular);
        Matcher matcher = pattern.matcher(string);
        return string;
    }

3 集合类库：

public List stringToNumber3(String string){
        List list = new ArrayList<>();
        Pattern pattern = Pattern.compile("^[0-9]$");
        Matcher matcher = pattern.matcher(string);
        String result = matcher.replaceAll("");
        for (int i = 0;i < result.length();i++){
            list.add(result.substring(i,i + 1));
        }
        return list;
    }

相同的思路：

public List stringToNumber4(String string){
        List list = new ArrayList<>();
        Pattern pattern = Pattern.compile("^[0-9]$");
        Matcher matcher = pattern.matcher(string);
        for (String string1 : matcher.replaceAll("^[0-9]$").split(",")){
            if (string1.length() > 0){
                list.add(string1);
            }
        }
        return list;
    }

从字符串文本中获得数字：

 public List getDigit(String text){
        List longs = new ArrayList<>();
        Pattern pattern = Pattern.compile("(\\d+)");
        Matcher matcher = pattern.matcher(text);
        while (matcher.find()){
            String find = matcher.group(1).toString();
            longs.add(Long.valueOf(find));
        }
        return longs;
    }

另附：

//判断一个字符串是否都为数字
    public boolean isDigit(String string){
        return string.matches("^[0-9]+$");
    }
    //判断一个字符串是否都为数字
    public boolean isDigit2(String string){
        Pattern pattern = Pattern.compile("^[0-9]+$");
        Matcher matcher = pattern.matcher(string);
        return matcher.matches();
    }
    //截取数字
    public String getNumbers(String string){
        Pattern pattern = Pattern.compile("^(\\d+)$");
        Matcher matcher = pattern.matcher(string);
        while (matcher.find()){
            return matcher.group(0);
        }
        return "";
    }
    //截取非数字
    public String splitNotNumber(String string){
        Pattern pattern = Pattern.compile("^(\\D+)$");
        Matcher matcher = pattern.matcher(string);
        while (matcher.find()){
            return matcher.group(0);
        }
        return "";
    }

二. 这是因为你没搞清楚java中int类型的范围,在java中int取值是从-2147483648到2147483647,2的31次方.而你的 String line3[1]= "8613719716 ";超过了这个最大的值,当然会抛NumberFormatException这个异常,说明不能转换类型.去掉一个能成功是因为他在范围内.