php - 如何从 mysqli 结果构建正确的 json？

Question

我正在尝试从我的 mysqli 结果构建一个 json 对象。我该怎么做。目前它不会创建一个 json 外观的对象。

这是我的代码:

$result = $dataConnection->prepare("SELECT id, artist, COUNT(artist) AS cnt FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$result->execute();
if($result->error)
{
die("That didn't work. I get this: " . $result->error);
}
$result->bind_result($id, $artist, $count);
$data = array();
while($result->fetch()){
$data[] = '{ id :'.$id.', artist :'.$artist.', count :'.$count.'}';
}
echo json_encode($data);
$dataConnection->close();

我想要一个像这样的数据对象:

{"id":"27","artist":"myArtist","count":"29"},....

Answer 1

$result = $dataConnection->query("SELECT id, artist, COUNT(artist) AS count FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$data = array();
while($row = $result->fetch_assoc()){
    $data[] = $row;
}
echo json_encode($data);

说实话，mysqli 是一个糟糕的 API，不能直接在应用程序代码中使用。

帮自己一个忙，至少使用 PDO

$result = $dataConnection->prepare("SELECT id, artist, COUNT(artist) AS count FROM {$databasePrefix}users GROUP BY artist ORDER BY cnt DESC LIMIT 0 , 30");
$result->execute();
echo json_encode($result->fetchAll());

与 mysqli 不同，它的方法始终有效。