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为什么不能调用函数里面的变量?

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Release: 2016-06-23 14:19:46
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<?php  //定义常量define("EntTime", "2012-08-01");define("EntTime2", "2012-08-31");define("Query_field", "品号");define("Operate", "包含");define("requirement", "WDZ");//将常量转换为变量$EntTime = EntTime;$EntTime2 = EntTime2;$Query_field = Query_field;$Operate = Operate;$requirement = requirement;//自定义函数function jhRepPd(){	GLOBAL $PUR,$MOC;	switch($Operate){		case "包含":			if($Query_field=="品号"){				$PUR = "PURTH.TH004 like'%".$requirement."%' AND ";			}			break;	}}//去除日期中的"-"$a_date = "PURTG.TG003 >='".str_replace("-","",$EntTime)."'";$b_date = "PURTG.TG003 <='".str_replace("-","",$EntTime2)."'";//判断变量是否为空if(!empty($EntTime) && !empty($EntTime2) && $requirement!==""){	$date = "(".$a_date." AND ".$b_date.") AND ";	jhRepPd();};//sql语句$sql = "SELECT * FROM TB where {$date}{$PUR}dbId in('1','2','3')";//打印SQL语句echo $sql;?>
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--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?SELECT * FROM TB where (PURTG.TG003 >='20120801' AND PURTG.TG003 <='20120831') AND dbId in('1','2','3')--正确的结果应该是:SELECT * FROM TB where (PURTG.TG003 >='20120801' AND PURTG.TG003 <='20120831') AND PURTH.TH004 like'%WDZ%' AND dbId in('1','2','3')
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回复讨论(解决方案)

GLOBAL不会,看楼下~

请指教

function jhRepPd(){
GLOBAL $PUR,$MOC;
switch( $Operate){
case "包含":
if( $Query_field=="品号"){
$PUR = "PURTH.TH004 like'%".$requirement."%' AND ";
}
break;
}
}
中,套红的是外部变量,没有声明。被视为局部变量,无值
所以变量 $PUR 并未赋值。当然就没有期望的结果了
应写作

function jhRepPd(){    GLOBAL $PUR,$MOC;    switch(Operate){ //既然定义有常量,为什么不用        case "包含":            if(Query_field=="品号"){ //这里也是                $PUR = "PURTH.TH004 like'%".requirement."%' AND ";//还有这里            }            break;    }}
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