我如何从mysqli结果构建正确的json

$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();

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

$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);

实话告诉你，mysql在应用程序代码中是一个糟糕的API。

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());

不像mysqli，它的方法总是有效的。

不要为将在

上调用json_encode的values数组构建json

代替:

$data[] = '{ id :'.$id.', artist :'.$artist.', count :'.$count.'}';

$data[] = array("id"=>$id, "artist"=>$artist, "count"=>$count);

如果你使用mysqli，这里有一个例子。我将它与javascript ajax调用结合使用。
输出如下所示:[{"field1":"23","field2":"abc"},{"field1":"24","field2":"xyz"}]

$mysqli = mysqli_connect('localhost','dbUser','dbPassword','dbName');
/* check connection */
if (mysqli_connect_errno()) {
  printf("Connect failed: %s'n", mysqli_connect_error());
  exit();
}
$query = "SELECT field FROM table LIMIT 10";
if ($result = mysqli_query($mysqli, $query)) {
  $out = array();
  while ($row = $result->fetch_assoc()) {
    $out[] = $row;
  }
  /* encode array as json and output it for the ajax script*/
  echo json_encode($out);
  /* free result set */
  mysqli_free_result($result);
  /* close connection*/
  $mysqli->close();
}
/* close connection*/
$mysqli->close();

创建一个包含所有行的数组，然后执行:

echo json_encode($array)