PHP循环中的json_encode()

我想在获取数据库结果的同时在while循环中使用json_encode().这是我的代码：

<?

$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);

$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");

while($row = sqlite_fetch_array($results)) {
  $data = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));

sqlite_close($database);

?>

这个的输出是

{“response”:”lastUserID lastUser lastXPos lastYPos”}

我希望它是……

{“response”:[“1 Alex 10 12”, “2 Fred 27 59”, “3 Tom 47 19”]}

等等

所以我希望json_encode()函数将所有用户放入数组而不是最后一个.我该怎么做？谢谢

解决方法:

尝试：

<?

$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);

$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");

$data = array();

while($row = sqlite_fetch_array($results)) {
  $data[] = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));

sqlite_close($database);

?>