Question

根据stackoverflow中给出的示例，我尝试了greybox popup，从我的php程序中基于if条件的函数调用。

它给出了如下错误：解析错误：语法错误，第456行/home/public_html/atst/dc-detail-qty.php中的意外'='

if if条件为true的php部分代码，greybox popup应该有效：

if($dcecrbal < 0)
{
$pop=100;
echo "<span class='sty1'>Negative Bal: ".$dcecrbal." (Total ECR: ".$ecrdtl." - Total DCs: ".$dcdtl.") </span>";
//echo "<script>child_open();</script>";

echo "<script>";
pathArr = window.location.pathname.split('/');
    path = window.location.protocol + "//" + window.location.host+"/";
    for (i=1;i<pathArr.length-1;i++) path += pathArr[i]+"/";

    GB_showCenter('ECR', path+'ecrframe-qty.php' , 800, 600);
echo "</script>";
}

Answer 1

应该是这样的

echo "<script>
      pathArr = window.location.pathname.split('/');
    path = window.location.protocol + '//' + window.location.host+'/';
    for (i=1;i<pathArr.length-1;i++) path += pathArr[i]+'/';

    GB_showCenter('ECR', path+'ecrframe-qty.php' , 800, 600)";
echo "</script>";

从给出错误的函数调用greybox弹出窗口

1 个答案: