在PHP中，SQLite3Result对象尚未正确初始化

Question

I have that piece of code in my php file (correctly working):

$db = new Database();
$stmt = $db->prepare("SELECT * FROM bills WHERE group_id=:gid");
$stmt->bindValue(":gid", $_SESSION['group_id'], SQLITE3_INTEGER);
if (($result = $stmt->execute()) === false) {
  echo "bad";
}

$bills = $result;
while ($bill = $bills->fetchArray()) {
  do stuff....
}

Then I try to put all database related code in function:

function getBillsByGID($gid) {
    $db = new Database();
    $stmt = $db->prepare("SELECT * FROM bills WHERE group_id=:gid");
    $stmt->bindValue(":gid", $gid, SQLITE3_INTEGER);

    if (($result = $stmt->execute()) === false) {
        return null;
    }   

    return $result;
}

And in the original file:

$bills = getBillsByGID($_SESSION['group_id']);
while ($bill = $bills->fetchArray()) {
       do stuff...
}

That gives me message: "Warning: SQLite3Result::fetchArray(): The SQLite3Result object has not been correctly initialised in line 61" (line with while($bill = $bills->fetchArray()))

var_dump($bills) after calling the function gives object(SQLite3Result)#10 (0) { }

So how do I make a function which will work properly?

Answer 1

I think you need to fetch data in your function and then return your variable

function getBillsByGID($gid) {
    $db = new Database();
    $stmt = $db->prepare("SELECT * FROM bills WHERE group_id=:gid");
    $stmt->bindValue(":gid", $gid, SQLITE3_INTEGER);

    if ($stmt->execute() === false) {
        return null;
    }   

    $array = array();
    while($data = $stmt->fetchArray())
    {
         $array[] = $data;
    }

    return $array;
}

Now assign the function to a variable and loop throw to get your array's data

$bills = getBillsByGID($_SESSION['group_id']);
foreach($bills as $data) {
     //do stuff with $data
}