Question

Using:
MySQL 5.1 + PHP 5.3.5

MySQL:

id in "table" is defined as: mediumint(6) zerofill not null

I get the expected result when using:

$mysqli->query("SELECT id FROM table WHERE id = 1");
while($row = $ret->fetch_array(MYSQLI_ASSOC)) $arr[] = $row;
>>> $arr[0]["id"] = 000001

But not when I use prepared statement:

$ret = $mysqli->prepare("SELECT id FROM table WHERE id = ?");
call_user_func_array(array($ret,"bind_param"),array("i",1));
$ret->execute();
$ret->store_result();
$meta = $ret->result_metadata();
$fields = $meta->fetch_fields();
$cols = array(); $data = array();
foreach($fields as $field) $cols[] = &$data[$field->name];
call_user_func_array(array($ret, 'bind_result'), $cols);
while($ret->fetch()) {
    $row = array();
    foreach($data as $key => $val) $row[$key] = $val;
    $arr[] = $row;
}
>>> $arr[0]["id"] = 1

When trying the prepared statement directly in the MySQL console, it shows as expected (it is not MySQL).

According to PHP mysqli_stmt::bind_result documentation I found this:

Depending on column types bound variables can silently change to the
corresponding PHP type.

I need to show the number with the trailing zeros WITHOUT having to do it in a later step. There are so many fields with zerofill and the process is practically automated from the data to the screen, so trying to fix this after this code, will require mayor changes.

Answer 1

Another solution was to specify the field as decimal(6,0). For some reason it works as expected (no changes to the code required).