java 将 String（代表十进制数）转换为 long
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Convert String (representing decimal number) to long
提问by temelm
I've googled around a bit but could not find examples to find a solution. Here is my problem:
我已经用谷歌搜索了一下，但找不到找到解决方案的例子。这是我的问题：
String s = "100.000";
long l = Long.parseLong(s);
The 2nd line of code which tries to parse the string 's' into a long throws a NumberFormatException.
尝试将字符串 's' 解析为 long 的第二行代码抛出NumberFormatException。
Is there a way around this? the problem is the string representing the decimal number is actually time in milliseconds so I cannot cast it to int because I lose precision.
有没有解决的办法？问题是代表十进制数的字符串实际上是以毫秒为单位的时间，所以我不能将它转换为 int，因为我失去了精度。
回答by assylias
You could use a BigDecimal to handle the double parsing (without the risk of precision loss that you might get with Double.parseDouble()):
您可以使用 BigDecimal 来处理双重解析（没有可能会导致精度损失的风险Double.parseDouble()）：
BigDecimal bd = new BigDecimal(s);
long value = bd.longValue();
回答by user902383
as i don't know is your 100.000 equals 100 or 100 000 i think safest solution which i can recommend you will be:
因为我不知道你的 100.000 是等于 100 还是 100 000 我认为我可以推荐你的最安全的解决方案是：
NumberFormat nf = NumberFormat.getInstance();
Number number = nf.parse("100.000");
long l = number.longValue();
回答by Kenogu Labz
'long' is an integer type, so the String parse is rejected due to the decimal point. Selecting a more appropriate type may help, such as a double.
'long' 是整数类型，因此字符串解析因小数点而被拒绝。选择更合适的类型可能会有所帮助，例如 double。
回答by Thomas
Just remove all spaceholders for the thousands, the dot...
只需删除数以千计的所有空格符，点...
s.replaceAll(".","");
回答by psabbate
You should use NumberFormat to parse the values
您应该使用 NumberFormat 来解析值
回答by jean joseph
We can use regular expression to trim out the decimal part. Then use parseLong
我们可以使用正则表达式来去掉小数部分。然后使用 parseLong
Long.parseLong( data.replaceAll("\..*", ""));
Long.parseLong( data.replaceAll("\..*", ""));
回答by Amit Deshpande
If you don't want to loose presision then you should use multiplication
如果你不想放松压力，那么你应该使用 multiplication
BigDecimal bigDecimal = new BigDecimal("100.111");
long l = (long) (bigDecimal.doubleValue() * 1000);<--Multiply by 1000 as it
is miliseconds
System.out.println(l);
Output:
输出：
100111
