为什么不能调用函数里面的变量？
//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;
?>–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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——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 “;
}