Php 如何检查mysqli中的错误？

我有下面的代码。。。有时候，这段代码只是告诉我一切正常，以及“成功”消息。但insert中的新行并没有添加到数据库中。 我想：echo$this->mysqli->error；将给我错误，但它不工作

if ($stmt = $this->mysqli->prepare("INSERT INTO tournaments ("
                    . "id_system, "
                    . "id_rank_admin, "
                    . "id_tournament_class, "
                    . "id_season, "
                    . "tournament_name,"
                    . "description,"
                    . "city,"
                    . "address,"
                    . "tournament_date,"
                    . "start_time,"
                    . "entry_fee,"
                    . "tournament_type,"
                    . "accepted_expansions,"
                    . "prices,"
                    . "additional_info,"
                    . "status,"
                    . "organizer_name,"
                    . "organizer_logo,"
                    . "organizer_link) VALUES (?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?,?)  ")) 
            {
                $stmt->bind_param("sssssssssssssssssss", 
                        $this->id_system, 
                        $this->id_rank_admin, 
                        $this->it_tournament_class,
                        $this->id_season, 
                        $this->tournament_name, 
                        $this->description, 
                        $this->city, 
                        $this->address, 
                        $this->tournament_date, 
                        $this->start_time, 
                        $this->entry_fee, 
                        $this->tournament_type, 
                        $this->accepted_expansions, 
                        $this->prices, 
                        $this->additional_info, 
                        $this->status, 
                        $this->organizer_name, 
                        $this->organizer_logo, 
                        $this->organizer_link);
                $stmt->execute();
                $stmt->close();
            }
            else
            {
                echo $this->mysqli->error;
            }

我在这里找到了这个：

返回最近一次MySQLi函数调用的最后一条错误消息，该调用可能成功，也可能失败

可能会让您更好地了解错误所在。

尝试以下方法：

$link = mysqli_connect("localhost", "my_user", "my_password", "db");

else
{
    echo(mysqli_error($link));
}

---

mysqli->execute what is is returns true or false在此基础上你可以得到你的错误我不太了解mysqli oop的方式，所以如果任何语法错误，请更正它

如果我没有错，使用mysqli只是不同风格的“程序化”而不是“面向对象”。。。

$link = mysqli_connect("localhost", "my_user", "my_password", "db");

else
{
    echo(mysqli_error($link));
}

$ok=$stmt->execute();

if($ok)
{
  echo "Query Executed Successfully";
}
else
{
 echo(mysqli_error($link));
}