CODEWITHSUNDEEP
phpmysql
user3583513
user3583513

Reputation:

Variable defined in script, but doesn't as parameter in function. what's wrong?

I'd like to have some class that has hold all about item that want to be sell, there are many item such as design, product, default product etc. and all product wanted to be hold in one database, so i can easily has a connection to table images, one to many.

so I created a class that would be hold all data in the item table... the class has to be know what kind of product it is. but the problem is when a certain column is undefined in function, but it is defined in script.

here is the MySQL table...

create table item (
    itemID int not null auto_increment,
    itemDiscount int default 0,
    orderCount int default 0,
    itemName text,
    designID int,
    embedID int,
    productID int,
    defaultProductID int,
    undesignID int,
    .
    .
    -- constraint declaration
);

I have inserted some data, here is the data (only for debugging purposes) 

itemID ... designID    productID    defaultProductID   undesignID   embedID
   1           1
   2                                       1
   3                        1
   4                                                                   1

class item{
    public $type;
    .
    .
    /**
     * Determine what kind of product in certain row.
     * @param array $row 
     * @return string
     */
    static function itemType($row){
        if(!is_null($row["productID"])){
            return "product";
        }else if(!is_null($row["defaultProductID"])){
            return "default";
        }else if(!is_null($row["designID"])){
            return "design";
        }else if(!is_null($row["embedID"])){
            $query  = "select * from embed where embedID={$row["embedID"]}";
            $result = execute($query);
            $row = mysqli_fetch_assoc($result);
            return "embed_" . self::itemType($row);
        }

    }
    function __construct(){
        //do stuff
    }
}

so i tried my static function :

$query = "select * from item";
$result = execute($query);
while($row = mysqli_fetch_assoc($result)){
     echo item::itemType($row) . "<br/>";
    //debug($row);
    echo $row["defaultProductID"] . "<br/>";
}

And this is I got:

design

default
1
product

Notice: Undefined index: defaultProductID in C:\Users\jodi\Documents\Visual Studio 2013\Projects\Prodegitv2\Prodegitv2\includes\item.php on line 17 

embed_design

Why I got that kind of notice?
1 is printed, so index defaultProductID has to be defined.

so I checked locals in debugging mode, here is the link for screenshot: http://postimg.org/image/ssfqv5f6h/

in visual studio debug mode, defaultProductID is defined...
Why does $row["defaultProductID"] be undefined in function??

is there any alternative solution??

Upvotes: 0

Views: 45

Answers (2)

Markus M&#252;ller
Markus M&#252;ller

Reputation: 2631

Your problem is with this code

    }else if(!is_null($row["embedID"])){
        $query  = "select * from embed where embedID={$row["embedID"]}";
        $result = execute($query);
        $row = mysqli_fetch_assoc($result);         <-----
        return "embed_" . self::itemType($row);     <-----
    }

You assign the query result of your query to the embed table to the $rowvariable overwriting what was in there before (the data of the item table). Then you call the itemType function again with your new $row data (from embed table). I guess there is no defaultProductId field in the embed table. This is the reason your code fails.

Upvotes: 2

Oli
Oli

Reputation: 2452

Try changing these lines:

$row = mysqli_fetch_assoc($result);
return "embed_" . self::itemType($row);

to: 

$res = mysqli_fetch_assoc($result);
return "embed_" . self::itemType($res);

Upvotes: 1

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