Can't see the problem in a simple php code!

[ashlegen]

Hi guys.I've got a problem.I'm building my php browsergame and i'm stuck into the inventory page.I have a code that seems ok to me but it won't work.Here it is:

 

 

$inventory = array();

$query = sprintf("SELECT id, item_id, quantity FROM user_items WHERE user_id = '%s'",

mysql_real_escape_string($userID));

$result = mysql_query($query);

while($row = mysql_fetch_assoc($result)) {

$item_query = sprintf("SELECT name FROM items WHERE id = '%s'",

mysql_real_escape_string($row['item_id']));

$item_result = mysql_query($item_query);

list($row['name']) = mysql_fetch_row($item_result);

array_push($inventory,$row);

}

 

 

 

I've spotted the problem.The $inventory array is empty.I checked it with if(empty($inventory)) .Thanks in advance and keep in mind that I'm a noob yet.

[mds1256]

Your SQL script is wrong, when using '%s' you need to use LIKE rather than =

[Pikachu2000]

No, that's a formatted string placeholder, not a wildcard operator.

 

Are you sure $userID has a valid value? Have you tried echo $query;?

[mds1256]

No, that's a formatted string placeholder, not a wildcard operator.

 

 

Sorry your correct, i missed the 2nd part of the line below

[Pikachu2000]

What exactly are you trying to make that code chunk do? Is it expected to return multiple records from the database?

[ashlegen]

Yes.I'm trying to get all of the item ids associated with a specific user id into an array.I'll keep you up to date and thanks for answering.

[ashlegen]

No, that's a formatted string placeholder, not a wildcard operator.

 

Are you sure $userID has a valid value? Have you tried echo $query;?

 

Oh man!I checked now and i've got the poblem.You were right.$uerID never had a value because i forgot the session_start() thing at the begiing of the document.Thanks again guys and I'm looking forward to ask more here.Cheers!