Alpha.Z写的数字转换为中文的类的补充 (我昨晚也是无聊)

发表于:2007-07-01来源:作者:点击数: 标签:
去掉了一些多余的零 ? php // 诸海加 2000-7-19 // xiaocui 2000-7-26 class ChineseNumber { var $basical=array(0=零,一,二,三,四,五,六,七,八,九); //var $basical=array(0=零,壹,贰,叁,肆,伍,陆,柒,捌,玖); var $advanced=array(1=十,百,千); //var $adva
去掉了一些多余的零

<?php
// 诸海加 2000-7-19
// xiaocui 2000-7-26

class ChineseNumber
{
    var $basical=array(0=>"零","一","二","三","四","五","六","七","八","九");
    //var $basical=array(0=>"零","壹","贰","叁","肆","伍","陆","柒","捌","玖");
    var $advanced=array(1=>"十","百","千");
    //var $advanced=array(1=>"拾","佰","仟");
    var $top=array(1=>"万","亿");

    var $level;            // 以4位为一级

    // 先实现万一下的数的转换
    function ParseNumber($number)
    {
        if ($number>999999999999)        // 只能处理到千亿。
            return "数字太大,无法处理。抱歉!";
        if ($number==0)
            return "零";

        for($this->level=0;$number>0.0001;$this->level++,$number=floor($number / 10000))
        {
            // 对于中文来说,应该是4位为一组。
            // 四个变量分别对应 个、十、百、千 位。
            $n1=substr($number,-1,1);
            if($number>9)
                $n2=substr($number,-2,1);
            else
                $n2=0;
            if($number>99)
                $n3=substr($number,-3,1);
            else
                $n3=0;
            if($number>999)
                $n4=substr($number,-4,1);
            else
                $n4=0;

            if($n4)
                $parsed[$this->level].=$this->basical[$n4].$this->advanced[3];
            else
                if(($number/10000)>=1)        // 千位为0,数值大于9999的情况
                    $parsed[$this->level].="零";
            if($n3)
                $parsed[$this->level].=$this->basical[$n3].$this->advanced[2];
            else
                if(!ereg("零$",$parsed[$this->level]) && ($number / 1000)>=1)    // 不出现连续两个“零”的情况
                    $parsed[$this->level].="零";
            if($n2)
                $parsed[$this->level].=$this->basical[$n2].$this->advanced[1];
            else
                if(!ereg("零$",$parsed[$this->level]) && ($number / 100)>=1)    // 不出现连续两个“零”的情况
                    $parsed[$this->level].="零";
            if($n1)
                $parsed[$this->level].=$this->basical[$n1];

            if($parsed[$this->level]!="零")
            {
               if(ereg("零$",$parsed[$this->level]))
                  $parsed[$this->level]=substr($parsed[$this->level],0,strlen($parsed[$this->level])-2);
               if($this->level>0)
                  $result=$parsed[$this->level].$this->top[$this->level].$result;
               else
                  $result=$parsed[$this->level].$result;
            }
        }
        //for($this->level-=1;$this->level>=0;$this->level--)
        //{
        //    $result.=$parsed[$this->level].$this->top[$this->level];
        //}

        //if(ereg("零$",$result))
        //    $result=substr($result,0,strlen($result)-2);

        return $result;
        
    }
};

$c=new ChineseNumber();
echo $c->ParseNumber(100000000001);
?>

原文转自:http://www.ltesting.net