我不太确定我是否理解您的增量案例以及您想要展示的内容。
不过,我决定提出一个格式化数字的解决方案。
我有两个版本的数字格式例程,一个解析数组,一个用正则表达式格式化。我承认它们不是最容易阅读的,但我很高兴想出这种方法。
我已经尝试用 cmets 来描述这些线条,以防你好奇
数组解析版本:
function formatNum(num) {
//Convert a formatted number to a normal number and split off any
//decimal places if they exist
var parts = String( num ).replace(/[^\d.]-/g,'').split('.');
//turn the string into a character array and reverse
var arr = parts[0].split('').reverse();
//initialize the return value
var str = '';
//As long as the array still has data to process (arr.length is
//anything but 0)
//Use a for loop so that it keeps count of the characters for me
for( var i = 0; arr.length; i++ ) {
//every 4th character that isn't a minus sign add a comma before
//we add the character
if( i && i%3 == 0 && arr[0] != '-' ) {
str = ',' + str ;
}
//add the character to the result
str = arr.shift() + str ;
}
//return the final result appending the previously split decimal place
//if necessary
return str + ( parts[1] ? '.'+parts[1] : '' );
}
正则表达式版本:
function formatNum(num) {
//Turn a formatted number into a normal number and separate the
//decimal places
var parts = String( num ).replace(/[^\d.]-/g,'').split('.');
//reverse the string
var str = parts[0].split('').reverse().join('');
//initialize the return value
var retVal = '';
//This gets complicated. As long as the previous result of the regular
//expression replace is NOT the same as the current replacement,
//keep replacing and adding commas.
while( retVal != (str = str.replace(/(\d{3})(\d{1,3})/,'$1,$2')) ) {
retVal = str;
}
//If there were decimal points return them back with the reversed string
if( parts[1] ) {
return retVal.split('').reverse().join('') + '.' + parts[1];
}
//return the reversed string
return retVal.split('').reverse().join('');
}
假设您希望每秒输出一个以 0.54 为增量的格式化数字,您可以使用间隔来进行增量和输出。
仅带有 Firebug 的超短 Firefox 示例:
var num = 1412015;
setInterval(function(){
//Your 0.54 value... why? I don't know... but I'll run with it.
num += 0.54;
console.log( formatNum( num ) );
},1000);
您可以在这里看到这一切:http://jsbin.com/opoze