在JavaScript中将长数字转换为缩写字符串,具有特殊的短缺要求 [英] Convert long number into abbreviated string in JavaScript, with a special shortness requirement
问题描述
在JavaScript中,如何编写一个函数将给定的[edit:正整数]数字(低于1000亿)转换为3个字母的缩写 - 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很小)不会,并且会在字母限制方面被忽略?
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?
这问题与有用线程,但它不一样;例如,该功能将转向例如123456 - > 1.23k(123.5k是5个字母)我正在寻找123456 - > 0.1m(0 [。] 1m为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!在做出以下修改时,答案在于并按照要求运作:
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) shortNum = shortValue.toFixed(1);
newValue = shortValue+suffixes[suffixNum];
}
return newValue;
}
推荐答案
我相信ninjagecko的解决方案没有'完全符合你想要的标准。以下函数:
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) {
var shortNum = shortValue.toFixed(1);
}
return shortValue+suffixes[suffixNum];
}
对于超过99万亿的值,不会添加任何字母,这很容易通过附加'后缀'数组来修复。
For values greater than 99 trillion no letter will be added, which can be easily fixed by appending to the 'suffixes' array.
由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) shortNum = shortValue.toFixed(1);
newValue = shortValue+suffixes[suffixNum];
}
return newValue;
}
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