我正在尝试将数字转换为文本,但是当我尝试将数字转换为未显示的十进制值但是只输入数字时输出的十进制值完美且十进制值为MAX 2位数我不确定那是怎么做的。
HTML:
mousePos
JavaScript的:
<input type="text" name="number" placeholder="Number OR Amount" onkeyup="word.innerHTML=convertNumberToWords(this.value)" />
<div id="word"></div>
答案 0 :(得分:1)
我找到了另一种解决方案。
var th = ['', 'thousand', 'million', 'billion', 'trillion'];
var dg = ['zero', 'one', 'two', 'three', 'four', 'five', 'six', 'seven', 'eight', 'nine'];
var tn = ['ten', 'eleven', 'twelve', 'thirteen', 'fourteen', 'fifteen', 'sixteen', 'seventeen', 'eighteen', 'nineteen'];
var tw = ['twenty', 'thirty', 'forty', 'fifty', 'sixty', 'seventy', 'eighty', 'ninety'];
function toWords(s) {
s = s.toString();
s = s.replace(/[\, ]/g, '');
if (s != parseFloat(s)) return 'not a number';
var x = s.indexOf('.');
if (x == -1) x = s.length;
if (x > 15) return 'too big';
var n = s.split('');
var str = '';
var sk = 0;
for (var i = 0; i < x; i++) {
if ((x - i) % 3 == 2) {
if (n[i] == '1') {
str += tn[Number(n[i + 1])] + ' ';
i++;
sk = 1;
} else if (n[i] != 0) {
str += tw[n[i] - 2] + ' ';
sk = 1;
}
} else if (n[i] != 0) {
str += dg[n[i]] + ' ';
if ((x - i) % 3 == 0) str += 'hundred ';
sk = 1;
}
if ((x - i) % 3 == 1) {
if (sk) str += th[(x - i - 1) / 3] + ' ';
sk = 0;
}
}
if (x != s.length) {
var y = s.length;
str += 'point ';
for (var i = x + 1; i < y; i++) str += dg[n[i]] + ' ';
}
return str.replace(/\s+/g, ' ');
}
答案 1 :(得分:1)
使用相同的函数获取整数和小数部分
function withDecimal(n) {
var nums = n.toString().split('.')
var whole = convertNumberToWords(nums[0])
if (nums.length == 2) {
var fraction = convertNumberToWords(nums[1])
return whole + 'and ' + fraction;
} else {
return whole;
}
}
console.log(withDecimal(51.32)) //Fifty One and Thirty Two
console.log(withDecimal(29.0)) //Twenty Nine
答案 2 :(得分:0)
我使用 Mr.isvforall 的想法对我的 eCheck 打印应用几乎没有增加。
function withDecimal(n) {
var nums = n.toString().split('.')
var whole = convertNumberToWords(nums[0])
if (nums.length == 2) {
var fraction = convertNumberToWords(nums[1])
// return whole + 'and ' + fraction;
return whole + 'and ' + nums[1] + "/100";
} else {
return whole;
}
}