在 JavaScript 中将长数字转换为缩写字符串，具有特殊的短要求

2022-01-17 00:00:00 numbers javascript

在 JavaScript 中，如何编写一个函数将给定的 [edit: positive integer] 数字(低于 1000 亿)转换为 3 字母缩写 - 其中 0-9 和 az/AZ 算作一个字母，但点(因为它在许多比例字体中是如此之小)不会，并且在字母限制方面会被忽略?

In JavaScript, how would one write a function that converts a given [edit: positive integer] number (below 100 billion) into a 3-letter abbreviation -- where 0-9 and a-z/A-Z are counting as a letter, but the dot (as it's so tiny in many proportional fonts) would not, and would be ignored in terms of the letter limit?

这个问题与 这个有用的线程，但不一样；例如，该功能将转向的地方，例如123456 -> 1.23k"(123.5k"是 5 个字母)我正在寻找可以执行123456 -> 0.1m"(0[.]1m"是 3 个字母)的东西.例如，这将是希望函数的输出(左原始，右理想返回值):

This question is related to this helpful thread, but it's not the same; for instance, where that function would turn e.g. "123456 -> 1.23k" ("123.5k" being 5 letters) I am looking for something that does "123456 -> 0.1m" ("0[.]1m" being 3 letters). For instance, this would be the output of hoped function (left original, right ideal return value):

0                      "0"
12                    "12"
123                  "123"
1234                "1.2k"
12345                "12k"
123456              "0.1m"
1234567             "1.2m"
12345678             "12m"
123456789           "0.1b"
1234567899          "1.2b"
12345678999          "12b"

谢谢！

更新:谢谢！当进行以下修改时，答案就在并按照要求工作:

function abbreviateNumber(value) {
    var newValue = value;
    if (value >= 1000) {
        var suffixes = ["", "k", "m", "b","t"];
        var suffixNum = Math.floor( (""+value).length/3 );
        var shortValue = '';
        for (var precision = 2; precision >= 1; precision--) {
            shortValue = parseFloat( (suffixNum != 0 ? (value / Math.pow(1000,suffixNum) ) : value).toPrecision(precision));
            var dotLessShortValue = (shortValue + '').replace(/[^a-zA-Z 0-9]+/g,'');
            if (dotLessShortValue.length <= 2) { break; }
        }
        if (shortValue % 1 != 0)  shortValue = shortValue.toFixed(1);
        newValue = shortValue+suffixes[suffixNum];
    }
    return newValue;
}

推荐答案

我相信 ninjagecko 的解决方案不太符合您想要的标准.以下函数可以:

I believe ninjagecko's solution doesn't quite conform with the standard you wanted. The following function does:

function intToString (value) {
    var suffixes = ["", "k", "m", "b","t"];
    var suffixNum = Math.floor((""+value).length/3);
    var shortValue = parseFloat((suffixNum != 0 ? (value / Math.pow(1000,suffixNum)) : value).toPrecision(2));
    if (shortValue % 1 != 0) {
        shortValue = shortValue.toFixed(1);
    }
    return shortValue+suffixes[suffixNum];
}

对于大于 99 万亿的值，不会添加任何字母，可以通过附加到后缀"数组轻松修复.

For values greater than 99 trillion no letter will be added, which can be easily fixed by appending to the 'suffixes' array.

Philipp 编辑如下:通过以下更改，它完全符合所有要求！

Edit by Philipp follows: With the following changes it fits with all requirements perfectly!

function abbreviateNumber(value) {
    var newValue = value;
    if (value >= 1000) {
        var suffixes = ["", "k", "m", "b","t"];
        var suffixNum = Math.floor( (""+value).length/3 );
        var shortValue = '';
        for (var precision = 2; precision >= 1; precision--) {
            shortValue = parseFloat( (suffixNum != 0 ? (value / Math.pow(1000,suffixNum) ) : value).toPrecision(precision));
            var dotLessShortValue = (shortValue + '').replace(/[^a-zA-Z 0-9]+/g,'');
            if (dotLessShortValue.length <= 2) { break; }
        }
        if (shortValue % 1 != 0)  shortValue = shortValue.toFixed(1);
        newValue = shortValue+suffixes[suffixNum];
    }
    return newValue;
}

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