getting PHP warning while running application

Question

I am getting a warning in php while running my application

HTML code:

      <a href="cart.html" class="dropdown-toggle" data-toggle="dropdown"><span> <?php if($a==true) { ?><?php echo $d->n;  ?> <?php } ?></span></a>

PHP code:

      $q=$mysqli->query("select object from object_store where Email='$email' ");
      $r=$q->fetch_assoc();
      $d=unserialize($r['object']);
      if(!is_null($d))
      {
      $a=true;
      }
      else
      {
      $a=false;
      }

The warning is cant call a member function on non object. I know that at first the value of $d is NULL but having used $a variable for check still the code block of $d is getting executed. Please suggest a way to remove this warning.

Most likely the warning references a specific error number? I would guess that command: $q->fetch_assoc(); since it is the only line in your code that actually uses any object. Actually this is a pretty typical error that has been answered endless times before: you do not do any error checking or handling at all. you blindly trust that your query well succeed. Most likely it does not succeed, but the call returns a boolean false. You cannot call a method on a boolean value. — arkascha
– arkascha, Commented Nov 22, 2015 at 18:02
my default object value is null and the query is running correctly — Sumeet Lalla
– Sumeet Lalla, Commented Nov 22, 2015 at 18:06
What is a "default object value"? And what line is referenced in the error you get? — arkascha
– arkascha, Commented Nov 22, 2015 at 18:06
in the HTML code which encloses $a boolean value it is displaying the warning — Sumeet Lalla
– Sumeet Lalla, Commented Nov 22, 2015 at 18:08

Berriel · Accepted Answer · 2015-11-22 18:09:46Z

1

I think this code:

$d=unserialize($r['object']);
if(!is_null($d))

will always be true, since unserialize doesn't return null.

Try: if($d != false) or if(is_array($d))

answered Nov 22, 2015 at 18:09

Berriel

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