Question

我正在努力完成三件事 -

我想缩短大数字并添加K / M / B后缀我希望能够强制小数位数我希望能够强制数千表示为百万的小数部分

缩短，舍入到2位小数

1200000 ----＆gt;＆gt;＆gt; 1.2M
1248000 ----＆gt;＆gt;＆gt; 1.25M
248000 ----＆gt;＆gt;＆gt; 248K

缩短，强制2位小数

1200000 ----＆gt;＆gt;＆gt; 1.20M
1248000 ----＆gt;＆gt;＆gt; 1.25M
248000 ----＆gt;＆gt;＆gt; 248.00K

缩短，强制3位小数，强制数千到数百万

1200000 ----＆gt;＆gt;＆gt; 1.200M
1248000 ----＆gt;＆gt;＆gt; 1.248M
248000 ----＆gt;＆gt;＆gt; 0.248M

我有一个javascript函数，我发现它做了很多，除了它不强制小数位数，它不允许我强制数千到数百万

function shortenNumber (num, decimalPlaces) {
var str,
    suffix = '';

decimalPlaces = decimalPlaces || 0;
num = +num;

var factor = Math.pow(10, decimalPlaces);


//99999 -> 99.9K

if (num < 1000) {
    str = num;
} else if (num < 1000000) {
    str = Math.floor(num / (1000 / factor)) / factor;
    suffix = 'K';
} else if (num < 1000000000) {
    str = Math.floor(num / (1000000 / factor)) / factor;
    suffix = 'M';
} else if (num < 1000000000000) {
    str = Math.floor(num / (1000000000 / factor)) / factor;
    suffix = 'B';
} else if (num < 1000000000000000) {
    str = Math.floor(num / (1000000000000 / factor)) / factor;
    suffix = 'T';
}
return str + suffix;
}

所以它完成了第一个要求，部分完成了第二个要求（它将舍入到2个小数位，但如果小数点为0则会丢弃它），但不能将K表示为M的

如何修改此功能以执行此操作（或将其替换为其他功能）？

谢谢！

Answer 1

这应该按照你的要求行事：

function abbreviate(number, maxPlaces, forcePlaces, forceLetter) {
  number = Number(number)
  forceLetter = forceLetter || false
  if(forceLetter !== false) {
    return annotate(number, maxPlaces, forcePlaces, forceLetter)
  }
  var abbr
  if(number >= 1e12) {
    abbr = 'T'
  }
  else if(number >= 1e9) {
    abbr = 'B'
  }
  else if(number >= 1e6) {
    abbr = 'M'
  }
  else if(number >= 1e3) {
    abbr = 'K'
  }
  else {
    abbr = ''
  }
  return annotate(number, maxPlaces, forcePlaces, abbr)
}

function annotate(number, maxPlaces, forcePlaces, abbr) {
  // set places to false to not round
  var rounded = 0
  switch(abbr) {
    case 'T':
      rounded = number / 1e12
      break
    case 'B':
      rounded = number / 1e9
      break
    case 'M':
      rounded = number / 1e6
      break
    case 'K':
      rounded = number / 1e3
      break
    case '':
      rounded = number
      break
  }
  if(maxPlaces !== false) {
    var test = new RegExp('\\.\\d{' + (maxPlaces + 1) + ',}$')
    if(test.test(('' + rounded))) {
      rounded = rounded.toFixed(maxPlaces)
    }
  }
  if(forcePlaces !== false) {
    rounded = Number(rounded).toFixed(forcePlaces)
  }
  return rounded + abbr
}

abbreviate(1200000, 2, false, false)
abbreviate(1248000, 2, false, false)
abbreviate(248000, 2, false, false)

abbreviate(1200000, 2, 2, false)
abbreviate(1248000, 2, 2, false)
abbreviate(248000, 2, 2, false)

abbreviate(1200000, 3, 3, 'M')
abbreviate(1248000, 3, 3, 'M')
abbreviate(248000, 3, 3, 'M')

Answer 2

如果你不反对外部图书馆，那就是Numeral.js。

javascript：缩短大数字，强制小数位，并选择将1000表示为.001百万

2 个答案: