Java 验证双精度数是否包含 2 个小数位的简单方法是什么？
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What is a simple way to validate if a double that contains 2 decimal places?
提问by user676567
I'm looking for an example in Java to check if a double amount entered by a user contains 2 decimal places e.g. 99.99 entered would return valid and 99.9 or 99.999 entered would return invalid.
我正在寻找一个 Java 示例来检查用户输入的双倍金额是否包含 2 个小数位，例如输入 99.99 将返回有效，而输入 99.9 或 99.999 将返回无效。
回答by Abimaran Kugathasan
Try the following regex
试试下面的正则表达式
\d+(\.\d{2})?
回答by Harsh Goswami
Double d = 234.12413;
String[] splitter = d.toString().split("\.");
splitter[0].length(); // Before Decimal Count
int decimalLength = splitter[1].length(); // After Decimal Count
if (decimalLength == 2)
// valid
else
// invalid
For Henry's question
对于亨利的问题
double d1 = 0.50;
double d2 = d1%1;
DecimalFormat df = new DecimalFormat("#.00");
int decimalLength = (df.format(d2).length()-1);
if (decimalLength == 2)
//valid
else
// invalid
Note : df.format(d2)returns .50, so the lengthis 3
注意：df.format(d2)返回.50，所以长度是3
回答by Ashot Karakhanyan
One more approach:
另一种方法：
//assert that the string is valid number
String num = "99.99"; // I assume that is `String` because entered by user and not yet converted to `double`
int i = num.lastIndexOf('.');
if(i != -1 && num.substring(i + 1).length() == 2) {
System.out.println("The number " + num + " has two digits after dot");
}
*ADDED
If you wand use local specific decimal separatoruse the result of following expession instead of '.'.
*添加如果你想使用本地特定的decimal separator使用以下表达式的结果而不是'.'.
DecimalFormat format = (DecimalFormat) DecimalFormat.getInstance();
DecimalFormatSymbols symbols = format.getDecimalFormatSymbols();
//symbols.getDecimalSeparator() == '.';
