3
A
回答
3
这个答案使用十进制格式化。它假设输入数字总是严格的形式( - )xx.yyy。
/**
* Converts a double of the form xx.yyy to xxyyy and -xx.yyy to -xxyy.
* No rounding is performed.
*
* @param number The double to format
* @return The formatted number string
*/
public static String format(double number){
DecimalFormat formatter = new DecimalFormat("#");
formatter.setRoundingMode(RoundingMode.DOWN);
number *= number < 0.0 ? 100 : 1000;
String result = formatter.format(number);
return result;
}
8
double yourDouble = 61.9155;
String str = String.valueOf(yourDouble).replace(".", "");
说明:
String.valueOf()
:转换你的双成Stringstr.replace(s1, s2)
:返回一个新字符串等于str
所有s1
小号
s2
取代'
更新:
的OP有一些额外的条件(但我不正好与一个知道):
负数 - >只有两位小数。
public static String doubleToSpecialString(double d) { if (d >= 0) { return String.valueOf(d).replace(".", ""); } else { return String.format("%.2f", d).replace(",", ""); } }
负数 - >一个小数较少
public static String doubleToSpecialString(double d) { if (d >= 0) { return String.valueOf(d).replace(".", ""); } else { String str = String.valueOf(d); int dotIndex = str.indexOf("."); int decimals = str.length() - dotIndex - 1; return String.format("%." + (decimals - 1) + "f", d).replace(",", ""); } }
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使用DecimalFormat而不是toString的+1使用 – 2010-09-09 17:47:23
没错!有用... – mtz 2010-09-10 07:35:38