java - 如何将包含 "e+"的字符串转换为长字符串？

Question

如1.3094266712875436e+18转换为1309426671287543600

我写了这样的方法，有更好的方法吗？

    public static void main(String[] args) {
        String a = "1.3094266712875436e+18";

        System.out.println(convert(a));

    }

    private static long convert(String a) {
        StringBuilder sb = new StringBuilder(a.length());
        StringBuilder sb2 = new StringBuilder(2);
        boolean isPlus = false;
        boolean countLength = false;
        int length = 0;
        for (char c : a.toCharArray()) {
            if ('.' == c) {
                countLength = true;
                continue;
            }
            if ('e' == c) {
                continue;
            }

            if ('+' == c) {
                isPlus = true;
                continue;
            }

            if (isPlus) {
                sb2.append(c);
            } else {
                sb.append(c);
                if (countLength) {
                    length++;
                }
            }
        }

        System.out.println(sb.toString());
        System.out.println(sb2.toString());

        long pow = 1;
        for (int i = 0; i < Integer.parseInt(sb2.toString()) - length; i++) {
            pow = pow * 10;
        }

        System.out.println(pow);

        return Long.parseLong(sb.toString()) * pow;
    }

打印出来

13094266712875436
18
100
1309426671287543600

Answer 1

您可以使用BigDecimal 。这将精确解析十进制字符串，然后您可以使用 longValue() 将其转换为 long。

String a = "1.3094266712875436e+18";
long value = new BigDecimal(a).longValue();
System.out.println(value);

输出:

1309426671287543600