Question

I have a simple table with one field “id”, and when I execute this code…

$dbh = new PDO('mysql:host='.$dbhost.';dbname='.$dbname, $dbuser, $dbpass);

$sql = 'SELECT * FROM logolist';
$q = $dbh->query($sql);
while($r = $q->fetch()){ print_r($r); }

… I get this output:

Array
(
    [ID] => 2
    [0] => 2
)
Array
(
    [ID] => 4
    [0] => 4
)

As you see, there’s a [0] under the field “ID”. if I add more field, I keep getting more extra elements inside the array. It’s like every field is outputting it’s value 2 times.

Why is this?

Answer 1

I’m encountering this practice of having a loop for fetching MySQL results and I’m wondering why people do it so I’ll write up this answer and try to clear up a few things.

1) You do not need a loop to fetch results
2) Reason you get the results duplicated is because you’re receiving an associative array and index-based one. That’s the default behaviour.

What you can do is this:

$dbh = new PDO('mysql:host='.$dbhost.';dbname='.$dbname, $dbuser, $dbpass);

// Tell PDO to throw exceptions in case of a query error
$dbh ->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

try
{
    $result = $dbh->query("SELECT * FROM logolist")->fetchAll(PDO::FETCH_ASSOC); 

    // Your result is now in $result array which is associative. 
    // If it's empty - no results were found. 
    // In case of an error, you'd go to catch block of the code and you'd echo out an error. 
}
catch(PDOException $e)
{
    echo "Error reported: ". $e->getMessage();
}