我在CodeIgniter中有这个查询：
$user_data = $this->db
->select('table_users.id AS user_id, table_users.email AS user_email, table_users.GUID as user_guid, '
. 'table_roles.slug AS role_slug, table_user_settings.username, table_users.id_roles, '
. 'table_users.first_name, table_users.last_name, table_users.mobile_number, table_users.phone_number, '
. 'table_users.address, table_users.city, table_users.state, table_users.zip_code, table_users.notes')
->from('table_users')
->where('table_users.data', 0)
->join('table_roles', 'table_roles.id = table_users.id_roles', 'inner')
->join('table_user_settings', 'table_user_settings.GUID = table_users.GUID')
->where('table_user_settings.username', $username)
->where('table_user_settings.password', $password)
->get()->row_array();
return ($user_data) ? $user_data : NULL;
我写了这段代码来转换它：
if($stmt = $this->db->prepare("SELECT table_users.id AS user_id, table_users.email AS user_email, table_users.GUID as user_guid, "
. "table_roles.slug AS role_slug, table_users.id_roles, table_users.first_name, "
. "table_users.last_name, table_users.mobile_number, table_users.phone_number, "
. "table_users.address, table_users.city, table_users.state, table_users.zip_code, table_users.notes "
. "FROM table_users "
. "INNER JOIN table_roles ON table_roles.id = table_users.id_roles "
. "INNER JOIN table_user_settings ON table_user_settings.GUID = table_users.GUID "
. "WHERE table_users.data = 0 AND "
. "table_user_settings.username = ? AND "
. "table_user_settings.password = ? "))
{
$stmt->bind_param("ss",$username, $password);
$stmt->bind_result($id, $email, $GUID, $slug, $id_roles, $first_name, $last_name,
$mobile_number, $phone_number, $address, $city, $state, $zip_code, $notes);
$result = $stmt->execute();
$stmt->fetch();
var_dump($stmt);
}
$stmt->close();
return $result;
但我不能返回一个数组中的所有变量作为第一个代码。在CodeIgniter中，我可以将所有结果作为数组获取，例如->get（）->row_array（）。但是我不能在 mysqli 中做同样的事情，我只想将查询的所有选定值作为数组返回......有人可以帮助我进行此转换吗？
如果我
理解正确，您想在不执行循环的情况下查询所有数组？
如果数据过多，查询可能需要一段时间，但您可以尝试直接从数据库中mysqli_fetch_all，如下所示（示例）：
$sql = $mysqli->query("SELECT * FROM table WHERE status = '1'");
$results = mysqli_fetch_all($sql, MYSQLI_ASSOC); # all rows to array
另外：你需要PHP>= 5.3.0mysqli_result：：fetch_all -- mysqli_fetch_all — 将所有结果行作为关联数组和/或数字数组读取
====
编辑：上面的方法避免了循环，但你需要自己转义这些值，如下所示：
$email = $mysqli->real_escape_string($email);
$GUID = $mysqli->real_escape_string($GUID);
要使用准备好的查询将所有内容全部发送到数组，您必须使用循环： 这是 mysqli 的一个限制，这就是为什么许多人更喜欢 pdo。因此，对于您的情况：
if (!$stmt->execute()) {
// handle error
}
// Extract result set and rows
$getresult = $stmt->get_result();
while ($data = $getresult->fetch_assoc()) {
$result[] = $data;
}
$stmt->close();
return $result;
// debug or testing
echo "<pre>";
var_dump($result);
echo "</pre>";