为啥不能调用函数里面的变量

为什么不能调用函数里面的变量？

<?php  <br />
<br />
//定义常量<br />
define("EntTime", "2012-08-01");<br />
define("EntTime2", "2012-08-31");<br />
define("Query_field", "品号");<br />
define("Operate", "包含");<br />
define("requirement", "WDZ");<br />
<br />
//将常量转换为变量<br />
$EntTime = EntTime;<br />
$EntTime2 = EntTime2;<br />
$Query_field = Query_field;<br />
$Operate = Operate;<br />
$requirement = requirement;<br />
<br />
//自定义函数<br />
function jhRepPd(){<br />
	GLOBAL $PUR,$MOC;<br />
	switch($Operate){<br />
		case "包含":<br />
			if($Query_field=="品号"){<br />
				$PUR = "PURTH.TH004 like'%".$requirement."%' AND ";<br />
			}<br />
			break;<br />
	}<br />
}<br />
<br />
//去除日期中的"-"<br />
$a_date = "PURTG.TG003 >='".str_replace("-","",$EntTime)."'";<br />
$b_date = "PURTG.TG003 <='".str_replace("-","",$EntTime2)."'";<br />
<br />
//判断变量是否为空<br />
if(!empty($EntTime) && !empty($EntTime2) && $requirement!==""){<br />
	$date = "(".$a_date." AND ".$b_date.") AND ";<br />
	jhRepPd();<br />
};<br />
<br />
//sql语句<br />
$sql = "SELECT * FROM TB where {$date}{$PUR}dbId in('1','2','3')";<br />
<br />
//打印SQL语句<br />
echo $sql;<br />
<br />
?>

--这是打印结果，但不是正确的。因为函数中的变量没有输出，为什么？<br />
SELECT * FROM TB where (PURTG.TG003 >='20120801' AND PURTG.TG003 <='20120831') AND dbId in('1','2','3')<br />
<br />
--正确的结果应该是：<br />
SELECT * FROM TB where (PURTG.TG003 >='20120801' AND PURTG.TG003 <='20120831') AND PURTH.TH004 like'%WDZ%' AND dbId in('1','2','3')

分享到：
------解决方案--------------------
function jhRepPd(){
    GLOBAL $PUR,$MOC;
    switch($Operate){
        case "包含":
            if($Query_field=="品号"){
                $PUR = "PURTH.TH004 like'%".$requirement."%' AND ";
            }