error php function on a non-object

Question

I have This php Function for get data for dynamic select box. i get country using form post method and put in dropdown menu. This function retrieve data from mysql :

   function drop_1($drop_var)
{  
    include_once('db.php');
    $result = mysql_query("SELECT * FROM home WHERE date='$drop_var' ") 
    or die(mysql_error());

    echo '<select name="drop_2" id="drop_2">
          <option value=" " disabled="disabled" selected="selected">Choose one</option>';

           while($drop_2 = mysql_fetch_array( $result )) 
            {
              echo '<option value="'.$drop_2['date'].'">'.$drop_2['data'].'</option>';
            }
 }

Now when i retireve data using jquery. I See this Error :

fatal error: Call to a member function query() on a non-object in C:\xampp\htdocs\test\func.php on line 14

This Line 14 : **$result = $db->query ($mysql);**

What's Problem ? Thanks

Where is $db defined? There error states that you try to acces query in the object $db. But that $db is not an object. So how do you define $db? — SynerCoder
– SynerCoder, Commented Mar 15, 2012 at 8:31
When you want to create object , first you need class for that , where is your class ? — Milap
– Milap, Commented Mar 15, 2012 at 8:37
@SynerCoder : i dont care ! whats your mean : $db = ........ — Saimon Avazian
– Saimon Avazian, Commented Mar 15, 2012 at 8:52
The whole point of your error is about how your defined $db. Is $db a object or not? Aka: is $db defined as followed: $db = new MyClass(); or $db = "string"; or $db = 12;. We need that line to discover your problem. — SynerCoder
– SynerCoder, Commented Mar 15, 2012 at 8:56

Saiyam Patel · Accepted Answer · 2012-03-15 08:33:03Z

1

$db belong to which class ? is there any class for $db object ? make that class OR include from some where...

answered Mar 15, 2012 at 8:33

Saiyam Patel

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3 Comments

Saimon Avazian Over a year ago

I Edited and Add $db function to my question

Saiyam Patel Over a year ago

$db = new DB... something like this OR directly use like $result = query($mysql);

Saimon Avazian Over a year ago

whitout db function this worked : include_once('db.php'); $result = mysql_query("SELECT * FROM data WHERE home='$drop_var'") or die(mysql_error()); while($drop_3 = mysql_fetch_array( $result ))

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