Question

我在下面有一个代码......有时这段代码只是告诉我一切正常，以及＆＃34;成功＆＃34;味精。但插入的新行未添加到数据库中。我想：echo $ this-＆gt; mysqli-＆gt;错误;会给我错误，但它不起作用？

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;
            }

Answer 1

$ok=$stmt->execute();

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

mysqli-＆gt;执行它所做的是基于它返回true或false你可以得到你的错误我没有太多的mysqli oop方式所以如果任何语法错误纠正它

Answer 2

我在这里找到了这个：http://php.net/manual/en/mysqli.error.php

 string mysqli_error ( mysqli $link )

返回最近可以成功或失败的MySQLi函数调用的最后一条错误消息。

可能会让您更好地了解错误的位置。

Answer 3

试试这个：

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

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

我如何检查mysqli中的错误？

3 个答案: