在AngularJS中格式化浮點數而不會損失精度

Question

在AngularJS中，如何在不丟失精度的情況下在HTML頁面上輸出浮點數，並且沒有不必要的填充0？

我已經考慮了“數字”ng-filter（ https://docs.angularjs.org/api/ng/filter/number ），但是fractionSize參數會產生固定的小數：

{{ number_expression | number : fractionSize}}

我正在尋找各種其他語言被稱為“精確再現性”，“規范字符串表示”，repr，往返等等，但我還沒有找到任何類似的AngularJS。

例如：

• 1 =>“1”
• 1.2 =>“1.2”
• 1.23456789 =>“1.23456789”

Answer 1

我自己偶然發現了一個明顯的解決方案！ 完全刪除“數字”ng-filter的使用將導致AngularJS根據我的要求簡單地將表達式轉換為字符串。

所以

{{ number_expression }}

代替

{{ number_expression | number : fractionSize}}

Answer 2

您可以捕獲零件而不尾隨零，並在正則表達式替換中使用它。 大概你想要保留一個尾隨零（例如“78.0”）以保持整潔而不是以小數分隔符結束（例如“78”）。

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"

regex101的解釋：

/([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