Question

在AngularJS中，如何在不丢失精度的情况下在HTML页面上输出浮点数，并且不使用0进行不必要的填充？

我考虑了“数字”ng-filter（https://docs.angularjs.org/api/ng/filter/number）但是fractionSize参数会产生固定的小数：

{{ number_expression | number : fractionSize}}

我正在寻找各种其他语言被称为“精确再现性”，“规范字符串表示”，repr，往返等等，但我无法找到类似AngularJS的任何内容。

例如：

1 =＆gt; “1”
1.2 =＆gt; “1.2”
1.23456789 =＆gt; “1.23456789”

Answer 1

我自己偶然发现了一个明显的解决方案！完全删除“数字”ng-filter的使用将导致AngularJS根据我的要求将表达式简单地转换为字符串。

所以

{{ number_expression }}

而不是

{{ number_expression | number : fractionSize}}

Answer 2

您可以在没有尾随零的情况下捕获零件，并在正则表达式替换中使用它。大概你想要保留一个尾随零（例如＆＃34; 78.0＆＃34;）以保持整洁，而不是以小数分隔符结束（例如＆＃34; 78。＆＃34;）。

var s = "12304.56780000";
// N.B. Check the decimal separator
var re = new RegExp("([0-9]+\.[0-9]+?)(0*)$");
var t = s.replace(re, '$1');  // t = "12304.5678"
t="12304.00".replace(re, "$1"); // t="12304.0"

来自regex101的解释：

/([0-9]+\.[0-9]+?)(0*)$/
    1st Capturing group ([0-9]+\.[0-9]+?)
        [0-9]+ match a single character present in the list below
            Quantifier: + Between one and unlimited times, as many times as possible, giving back as needed [greedy]
            0-9 a single character in the range between 0 and 9
        \. matches the character . literally
        [0-9]+? match a single character present in the list below
            Quantifier: +? Between one and unlimited times, as few times as possible, expanding as needed [lazy]
            0-9 a single character in the range between 0 and 9
    2nd Capturing group (0*)
        0* matches the character 0 literally
            Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
    $ assert position at end of the string

在AngularJS中格式化浮点数而不会损失精度

2 个答案: