DROP和CREATE在PHP中查看 - 单个查询

Question

我正在尝试使用PHP中的代码在mySQL中删除并创建一个视图：

$hours_sum_build = "DROP VIEW hours_sum;";
$hours_sum_build .= "CREATE VIEW hours_sum as
  SELECT emp_id, first_name, last_name, pp_end, date, sum(hours) as 
  hours, time_source, run_date, run_time
  FROM merged_time_report GROUP BY emp_id";

mysqli_multi_query($conn, $hours_sum_build);

echo "Debug: $hours_sum_build";
$result = mysqli_multi_query($conn, $hours_sum_build);
if ( false===$result ) {
  printf("error: %s\n", mysqli_error($conn));
}
 else {
 echo 'done.';
}

我显然希望DROP视图然后立即再创建它。但是，当我运行上面的代码时，我收到此错误消息：

Debug: DROP VIEW hours_sum;CREATE VIEW hours_sum as
SELECT emp_id, first_name, last_name, pp_end, date, sum(hours) as 
hours, time_source, run_date, run_time
FROM merged_time_report GROUP BY emp_id

error: 

Commands out of sync; you can't run this command now
null results for hours_sum_count
null results for first_pass_count
null results for pass_fail_count
Error: 
Commands out of sync; you can't run this command now
Error: 
Commands out of sync; you can't run this command now
Error: 
Commands out of sync; you can't run this command now
null

创建视图时（我可以通过phpMyAdmin完成），它会返回上面标识的三个变量（hours_sum_count等）中的结果，因此在我的场景中这些是空的，因为视图不是使用mysqli_multi_query在PHP中创建。

感谢您的帮助！