插入逗号到数字字符串 [英] Insert commas into number string

问题描述

嘿，我想在一个字符串进行向后定期EX pression搜索，将其划分成3位数的组。据我从AS3 <一见href="http://livedocs.adobe.com/flex/3/html/help.html?content=12%5FUsing%5FRegular%5FEx$p$pssions%5F01.html">documentation,反向搜索是不可能的了reg前引擎。

本练习的要点是插入三重逗号为若干个像这样：

 千万=＆GT; 10,000,000
 

我想这样做，像这样的：

 与string.replace（/（\ d {3}）/克，$ 1）
 

但是，这是不正确的，由于搜索没有从后面发生的事情和替换$ 1将只对第一场比赛。

我越来越觉得我会过得更好使用循环执行此任务。

更新：

由于AS3不支持预测先行这是我如何解决它。

 公共静态函数formatNumber（号码：号码）：字符串
{
VAR numString：字符串= number.toString（）
VAR的结果：string =''

而（numString.length→3）
{
VAR块：字符串= numString.substr（-3）
numString = numString.substr（0，numString.length  -  3）
结果='，'+大块+结果
}

如果（numString.length大于0）
{
结果= numString +结果
}

返回结果
}
 

解决方案

如果你的语言支持阳性向前断言，那么我想下面的正则表达式将工作：

 （\ d）（？=（\ d {3}）+ $）
 

展示在Java中：

 进口静电org.junit.Assert.assertEquals;

进口org.junit.Test;

公共类CommifyTest {

    @测试
    公共无效testCommify（）{
        串NUM0 =1;
        串NUM1 =123456;
        字符串NUM2 =1234567;
        串NUM3 =12345678;
        字符串num4 =123456789;

        字符串的正则表达式=（\\ D）（=（\\Ð{3}）+ $？）;

        的assertEquals（1，num0.replaceAll（正则表达式，$ 1，））;
        的assertEquals（123456，num1.replaceAll（正则表达式，$ 1））;
        的assertEquals（1,234,567，num2.replaceAll（正则表达式，$ 1））;
        的assertEquals（12345678，num3.replaceAll（正则表达式，$ 1））;
        的assertEquals（123,456,789，num4.replaceAll（正则表达式，$ 1））;
    }
}
 

下面的<一个href="http://livedocs.adobe.com/flex/3/html/help.html?content=12%5FUsing%5FRegular%5FEx$p$pssions%5F09.html#123006">link建议AS3呢？

Hey there, I'm trying to perform a backwards regular expression search on a string to divide it into groups of 3 digits. As far as i can see from the AS3 documentation, searching backwards is not possible in the reg ex engine.

The point of this exercise is to insert triplet commas into a number like so:

10000000 => 10,000,000

I'm thinking of doing it like so:

string.replace(/(\d{3})/g, ",$1")

But this is not correct due to the search not happening from the back and the replace $1 will only work for the first match.

I'm getting the feeling I would be better off performing this task using a loop.

UPDATE:

Due to AS3 not supporting lookahead this is how I have solved it.

public static function formatNumber(number:Number):String
{
	var numString:String = number.toString()
	var result:String = ''

	while (numString.length > 3)
	{
		var chunk:String = numString.substr(-3)
		numString = numString.substr(0, numString.length - 3)
		result = ',' + chunk + result
	}

	if (numString.length > 0)
	{
		result = numString + result
	}

	return result
}

解决方案

If your language supports postive lookahead assertions, then I think the following regex will work:

(\d)(?=(\d{3})+$)

Demonstrated in Java:

import static org.junit.Assert.assertEquals;

import org.junit.Test;

public class CommifyTest {

    @Test
    public void testCommify() {
        String num0 = "1";
        String num1 = "123456";
        String num2 = "1234567";
        String num3 = "12345678";
        String num4 = "123456789";

        String regex = "(\\d)(?=(\\d{3})+$)";

        assertEquals("1", num0.replaceAll(regex, "$1,"));
        assertEquals("123,456", num1.replaceAll(regex, "$1,"));
        assertEquals("1,234,567", num2.replaceAll(regex, "$1,"));
        assertEquals("12,345,678", num3.replaceAll(regex, "$1,"));
        assertEquals("123,456,789", num4.replaceAll(regex, "$1,"));    
    }    
}

The following link suggests AS3 does?

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