我首先使用了matti-lyra和gary的答案的组合:

r=(+n).toFixed(4).replace(/\.0+$/,'')

结果:

1234870.98762341:“1234870.9876” 1230009100:“1230009100” 0.0012234:“0.0012” 0.1200234:“0.12” 0.000001231:“0” 0.10001:“0.1000” "asdf": "NaN"(所以没有运行时错误)

有点问题的情况是0.10001。我最终使用了这个更长的版本:

    r = (+n).toFixed(4);
    if (r.match(/\./)) {
      r = r.replace(/\.?0+$/, '');
    }

1234870.98762341:“1234870.9876” 1230009100:“1230009100” 0.0012234:“0.0012” 0.1200234:“0.12” 0.000001231:“0” 0.10001:“0.1” "asdf": "NaN"(所以没有运行时错误)

更新:这是加里的新版本(见评论):

r=(+n).toFixed(4).replace(/([0-9]+(\.[0-9]+[1-9])?)(\.?0+$)/,'$1')

这将得到与上面相同的结果。

我首先使用了matti-lyra和gary的答案的组合:

r=(+n).toFixed(4).replace(/\.0+$/,'')

结果:

1234870.98762341:“1234870.9876” 1230009100:“1230009100” 0.0012234:“0.0012” 0.1200234:“0.12” 0.000001231:“0” 0.10001:“0.1000” "asdf": "NaN"(所以没有运行时错误)

有点问题的情况是0.10001。我最终使用了这个更长的版本:

    r = (+n).toFixed(4);
    if (r.match(/\./)) {
      r = r.replace(/\.?0+$/, '');
    }

1234870.98762341:“1234870.9876” 1230009100:“1230009100” 0.0012234:“0.0012” 0.1200234:“0.12” 0.000001231:“0” 0.10001:“0.1” "asdf": "NaN"(所以没有运行时错误)

更新:这是加里的新版本(见评论):

r=(+n).toFixed(4).replace(/([0-9]+(\.[0-9]+[1-9])?)(\.?0+$)/,'$1')

这将得到与上面相同的结果。

纯正则表达式的答案

n.replace(/(\.[0-9]*[1-9])0+$|\.0*$/,'$1');

我想知道为什么没有人给我一个!

如果将它转换为字符串，它将不会显示任何尾随零，因为它是作为数字而不是字符串创建的，所以后面的零就不会存储在变量中。

var n = 1.245000
var noZeroes = n.toString() // "1.245" 

如果由于任何原因(比如涉及到金钱浮动)不能使用float，并且已经从表示正确数字的字符串开始，您可以发现这个解决方案很方便。它将表示数字的字符串转换为表示数字的字符串，后面不带零。

function removeTrailingZeroes( strAmount ) {
    // remove all trailing zeroes in the decimal part
    var strDecSepCd = '.'; // decimal separator
    var iDSPosition = strAmount.indexOf( strDecSepCd ); // decimal separator positions
    if ( iDSPosition !== -1 ) {
        var strDecPart = strAmount.substr( iDSPosition ); // including the decimal separator

        var i = strDecPart.length - 1;
        for ( ; i >= 0 ; i-- ) {
            if ( strDecPart.charAt(i) !== '0') {
                break;
            }
        }

        if ( i=== 0 ) {
            return strAmount.substring(0, iDSPosition);
        } else {
            // return INTPART and DS + DECPART including the rightmost significant number
            return strAmount.substring(0, iDSPosition) + strDecPart.substring(0,i + 1);
        }
    }

    return strAmount;
}

如果我们有一个数字的s字符串表示形式，例如我们可以使用number的.toFixed(digits)方法(或任何其他方法)来获得，那么为了从s字符串中删除不重要的末尾零，我们可以使用:

s.replace(/(\.0*|(?<=(\..*))0*)$/, '')

/**********************************
 * Results for various values of s:
 **********************************
 *
 * "0" => 0
 * "0.000" => 0
 * 
 * "10" => 10
 * "100" => 100
 * 
 * "0.100" => 0.1
 * "0.010" => 0.01
 * 
 * "1.101" => 1.101
 * "1.100" => 1.1
 * "1.100010" => 1.10001
 * 
 * "100.11" => 100.11
 * "100.10" => 100.1
 */

replace()中使用的正则表达式解释如下:

In the first place please pay the attention to the | operator inside the regular expression, which stands for "OR", so, the replace() method will remove from s two possible kinds of substring, matched either by the (\.0*)$ part OR by the ((?<=(\..*))0*)$ part. The (\.0*)$ part of regex matches a dot symbol followed by all the zeros and nothing else till to the end of the s. This might be for example 0.0 (.0 is matched & removed), 1.0 (.0 is matched & removed), 0.000 (.000 is matched & removed) or any similar string with all the zeros after the dot, so, all the trailing zeros and the dot itself will be removed if this part of regex will match. The ((?<=(\..*))0*)$ part matches only the trailing zeros (which are located after a dot symbol followed by any number of any symbol before start of the consecutive trailing zeros). This might be for example 0.100 (trailing 00 is matched & removed), 0.010 (last 0 is matched & removed, note that 0.01 part do NOT get matched at all thanks to the "Positive Lookbehind Assertion", i.e. (?<=(\..*)), which is in front of 0* in this part of regex), 1.100010 (last 0 is matched & removed), etc. If neither of the two parts of expression will match, nothing gets removed. This might be for example 100 or 100.11, etc. So, if an input does not have any trailing zeros then it stays unchanged.

更多使用.toFixed(数字)的例子(在下面的例子中使用了字面值“1000.1010”，但我们可以假设变量):

let digits = 0; // Get `digits` from somewhere, for example: user input, some sort of config, etc.

(+"1000.1010").toFixed(digits).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000'

(+"1000.1010").toFixed(digits = 1).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000.1'


(+"1000.1010").toFixed(digits = 2).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000.1'


(+"1000.1010").toFixed(digits = 3).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000.101'


(+"1000.1010").toFixed(digits = 4).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000.101'


(+"1000.1010").toFixed(digits = 5).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000.101'

(+"1000.1010").toFixed(digits = 10).replace(/(\.0*|(?<=(\..*))0*)$/, '');
// Result: '1000.101'

要使用replace()中使用的上述正则表达式，我们可以访问:https://regex101.com/r/owj9fz/1