CODEWITHSUNDEEP
phpmysqlsql
LUX
LUX

Reputation: 97

SQL won't return values

Let me first start to say my SQL is correct and checked it more than once.

I have 2 methods in my file calling more than once for the SQL but when it comes to the second part it won't return any value to fill in the fields.

Here is the part that is giving me issues:

$query = mysql_query("SELECT * FROM MegaFilter WHERE Parent = '".$ThisPage."' ");
while($query2 = mysql_fetch_assoc($query)) {
    $SubID = $query2['Filter'];
    $query3 = mysql_query("SELECT * FROM SubCategories WHERE ID = '".$SubID."' ");
    while($query4 = mysql_fetch_assoc($query3)) {
        $SubCatID = $query4['ID'];
        $query5 = mysql_query("SELECT * FROM Web_Pages WHERE SubCat = '".$SubCatID."' ");
    }
    while($query6 = mysql_fetch_assoc($query5)) {
        $ProductName = $query6['Title'];
        $ProductID = $query6['ID'];
    }
    echo '<li class="item" data-id="id-'.$Productid.'" data-type="'.$SubID.'">'.$Productname.'</li>';
}

It does not log any errors except that the last 2 variables are not defined.

Upvotes: 1

Views: 135

Answers (2)

Ed Heal
Ed Heal

Reputation: 60027

  1. Should not be using the mysql_ libraries any more - they are deprecated.
  2. Your braces are all over the place (if I understand what you are trying to achieve see below).
  3. PHP variables are case sensitive.

I think that the code that you require is (just need the one select statement):

$connection = new mysqli('localhost', 'my_user', 'my_password', 'my_db'); 
$stmt = $connection -> mysqli_prepare('SELECT m.Filter AS SubID, w.Title AS ProductName, w.ID AS ProductID FROM MegaFilter AS m, INNER JOIN SubCategories AS sc ON (sc.ID = m.Filter) INNER JOIN Web_Pages AS w ON (w.SubCat = sc.ID) WHERE Parent = ?');
$stmt->bind_param('s', $ThisPage);
$stmt->bind_result($SubID, $ProductName, $ProductID);
$stmt->execute();
while ($stmt->fetch())
{
    echo '<li class="item" data-id="id-'.$ProductID.'" data-type="'.$SubID.'">'.$ProductName.'</li>'
}

Upvotes: 0

Aiias
Aiias

Reputation: 4748

Variable names are case sensitive in php. You are assigning values to $ProductName and $ProductID but you are using $Productid (lowercase i) and $Productname (lowercase n).

Upvotes: 1

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