CODEWITHSUNDEEP
phpfunctionpopupgreybox
user1114409
user1114409

Reputation: 555

Calling greybox popup from a function giving error

As per the example given in stackoverflow I tried greybox popup, calling from a function based on a if condition in my php program.

It gives an error like: Parse error: syntax error, unexpected '=' in /home/public_html/atst/dc-detail-qty.php on line 456

The php partial code with if condition true, the greybox popup should work:

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>";
}

Upvotes: 1

Views: 345

Answers (1)

Bhuvan Rikka
Bhuvan Rikka

Reputation: 2703

That should be like this

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>";

Upvotes: 0

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