php – 为什么我收到错误“命令不同步;你现在不能运行这个命令“

发布时间：2020-12-13 16:36:14 所属栏目：PHP教程来源：网络整理

导读：标题中提到的错误文档说明 If you get Commands out of sync; you can’t run this command now in your client code,you are calling client functions in the wrong order. This can happen,for example,if you are using mysql_use_result() and try to e

标题中提到的错误文档说明

If you get Commands out of sync; you can’t run this command now in
your client code,you are calling client functions in the wrong order.

This can happen,for example,if you are using mysql_use_result() and
try to execute a new query before you have called mysql_free_result().
It can also happen if you try to execute two queries that return data
without calling mysql_use_result() or mysql_store_result() in between.

从这里：http://dev.mysql.com/doc/refman/5.0/en/commands-out-of-sync.html

但是在First Query中我没有从mysql数据库中获取任何数据,我只是插入.在第二个查询中,我从数据库中获取数据.

这是我的代码

$connection = mysqli_connect("localhost","username","password","tbl_msgs");
if(mysqli_connect_errno($connection))
{
    die("Failed to connect to MySQL: " . mysqli_connect_error());
}
$query = "INSERT INTO users (total_comments,total_views) 
          VALUES ({$total_comments},{$total_views});";

$query .= "INSERT INTO msgs (notifications) VALUES ({$notifications})";

mysqli_multi_query($connection,$query);

每一件事情都很好.但是当我执行以下查询时它会给出错误

$select_query = "SELECT * FROM msgs WHERE msg_id = {$msg_id}";

$result_set = mysqli_query($connection,$select_query);

if(!$result_set) {
    die(mysqli_error($connection)); 
}

这里它使错误命令不同步;你现在不能运行这个命令.我无法理解这种情况

注意：查询中有任何问题,我已经直接向PHPMyAdmin执行了相同的查询,并且工作正常.

查询中有待处理的结果集：

mysqli_multi_query($connection,$query);

您需要先使用/存储结果,然后才能继续下一个查询：
由于您看起来并不真正关心第一个结果集,请在多个查询之后执行此操作.

do
{
    $result = mysqli_store_result($connection);
    mysqli_free_result($result);
}while(mysqli_next_result());

另一种方法是关闭连接并再次启动它.

mysqli_close($connection);
$connection = mysqli_connect("localhost","tbl_msgs");

这一切都取决于您的要求.

（编辑：李大同）

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