你可以试试这个来缩小浮点数

var n = 0.0000;
n = parseFloat(n.toString()); 

//output n = 0; 
// n = 3.14000; --> n = 3.14;

当Django在文本字段中显示十进制类型的值时，我也需要解决这个问题。例如，当'1'是值时。它会显示“1.00000000”。如果'1.23'是值，它将显示'1.23000000'(在'decimal_places'设置为8的情况下)

使用parseFloat对我来说不是一个选项，因为它可能不会返回完全相同的值。toFixed不是一个选项，因为我不想四舍五入任何东西，所以我创建了一个函数:

function removeTrailingZeros(value) {
    value = value.toString();

    # if not containing a dot, we do not need to do anything
    if (value.indexOf('.') === -1) {
        return value;
    }

    # as long as the last character is a 0 or a dot, remove it
    while((value.slice(-1) === '0' || value.slice(-1) === '.') && value.indexOf('.') !== -1) {
        value = value.substr(0, value.length - 1);
    }
    return value;
}

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

所以你想

var x = 1.234000; // to become 1.234
var y = 1.234001; // stays 1.234001

没有附加条件，只需尝试Number()。

Var x = 1.234000， //变成1.234 Y = 1.234001， //保持1.234001 x_ =数字(x)， y_ =数字(y); y_ console.log(间);

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" 

我认为下面的函数可能接近你想要的。我是为我的一个应用程序写的。它总是以不带后面零的标准符号输出。有些东西你可能不想要，但如果你喜欢，可以删掉。它总是返回至少一个小数(e.x。5 = >“5.0”)。它也被限制为10个小数。把它作为一个指南。

const toDecimalStr(value)=>{
  let str=value.toFixed(10).replace(/([0]+)$/,"");
  try {
    if (str.endsWith(".")) str+='0';
  } catch (e) {
    str+='0';
  }
  return str;
}

try catch是因为不是所有东西都支持endsWith，我太懒了。