java 从字符串中删除前导/尾随零的正则表达式
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Regular expression to remove leading/trailing zeros from a string
提问by Sam
Is this the proper REGEX to remove trailing decimal and zeroes from a string? I can't get it to work. What am I missing?
这是从字符串中删除尾随小数和零的正确正则表达式吗？我无法让它工作。我错过了什么？
78.80 -> 78.8
78.80 -> 78.8
str.replaceAll("^.0*$", "");
i need only 2 decimal points as well like 78.008 should be 78.01
我只需要 2 个小数点，就像 78.008 应该是 78.01
and if it is 78.10 the 78.1 only.
如果是 78.10，则只有 78.1。
回答by xanatos
No, the correct regular expression is something more complex, and it needs positive look-behind.
不，正确的正则表达式更复杂，它需要积极的回顾。
str = str.replaceAll("\.0*$|(?<=\.[0-9]{0,2147483646})0*$", "");
you have to escape the ., because otherwise it means "any character", you have to anchor the regex only to the end of the string, and you have to say that the 0digits must be after a .plus some optional non-zero digits.
你必须转义.，否则它意味着“任何字符”，你必须将正则表达式锚定到字符串的末尾，并且你必须说0数字必须在.加上一些可选的非零数字之后。
Addendum: there is a "special case" that should be handled: 1.00. We handle it by using the |. The first sub-expression means "a dot plus only zeroes" and matches even the dot (that in this way is deleted)
附录：有一个“特殊情况”应当处理：1.00。我们通过使用|. 第一个子表达式的意思是“一个点加上只有零”，甚至匹配这个点（以这种方式被删除）
And remember that in Java strings are immutable, so replaceAllwill create a new string.
请记住，在 Java 中字符串是不可变的，因此replaceAll会创建一个新字符串。
Note the use of {0,2147483646}: if you used a *you would get a Look-behind group does not have an obvious maximum length, because the *would be converted to {0,2147483647}and considering the "+1" length of the \\.it would overflow, so we put the maximum number of digits possible (2147483647, maximum value of a signed int) minus 1 for the dot.
注意使用{0,2147483646}: 如果你使用 a*你会得到 a Look-behind group does not have an obvious maximum length，因为*将被转换为{0,2147483647}并考虑到它的“+1”长度\\.会溢出，所以我们把可能的最大位数（2147483647，a的最大值有符号 int) 减 1 表示点。
Test example: http://ideone.com/0NDTSq
测试示例：http: //ideone.com/0NDTSq
回答by yiannis
Since you want rounding of the numbers, you should not use regular expressions. You should use DecimalFormat:
由于您想要对数字进行四舍五入，因此不应使用正则表达式。你应该使用DecimalFormat：
DecimalFormat twoDForm = new DecimalFormat("#.#");
Double d = Double.parseDouble("123.078");
String s = twoDForm.format(d);
System.out.println(s);
will output:
将输出：
123.08
回答by stema
With the anchors ^and $you are trying to match the whole string and replace it. And this will not match since there is more in the string.
使用锚点^，$您正在尝试匹配整个字符串并替换它。这将不匹配，因为字符串中有更多内容。
Try this
试试这个
str.replace("0*$", "");
With 0*$you will match only 0 or more "0" at the end of the string.
随着0*$你只会0或多个“0”的字符串的结尾相匹配。
Note:This will also change "1000" to "1"!
注意：这也会将“1000”更改为“1”！
