Question

我有一个连接类，它成功连接到所需的数据库表。我已经执行了插入查询，我想获取连接的最后插入的ID。但是我收到一个Notice: Undefined property: connection::$insert_id错误。我在连接对象上var_dump($con)，下面是我得到的。希望有人可以帮助我解决这个问题。非常感谢您的帮助。

对象（连接）＃2（5）{[“” db_name“] =>字符串（5）” dbproject“ [” db_user“] =>字符串（4）” root“ [” db_pass“] =>字符串（ 0）“” [“ db_host”] =>字符串（9）“ localhost” [“ connection”] => object（mysqli）＃3（19）{[“ affected_rows”] => int（1）[“ client_info” ] =>字符串（79）“ mysqlnd 5.0.12-dev-20150407-$ Id：241ae00989d1995ffcbbf63d579943635faf9972 $” [“ client_version”] => int（50012）[“ connect_errno”] => int（0）[“ connect_error”] => NULL [“ errno”] => int（0）[“ error”] =>字符串（0）“” [“ error_list”] => array（0）{} [“ field_count”] => int（0 ）[“ host_info”] =>字符串（20）“通过TCP / IP的本地主机” [“ info”] => NULL [“ insert_id”] => int（26）[“ server_info”] =>字符串（21）“ 5.5.5-10.1.19-MariaDB“ [” server_version“] => int（50505）[” stat“] => string（136）”正常运行时间：302248线程：1问题：9631慢查询：0打开：65刷新表：1个打开的表：40个每秒的查询平均值：0.031“ [” sqlstate“] => string（5）” 00000“ [” protocol_version“] => int（10）[” thread_id“] => int（522） [“ warning_count”] => int（0）}}

class connection {

    public $db_name;
    public $db_user;
    public $db_pass;
    public $db_host;

    public $connection;

     public function connection(){
        $this->db_host = HOST;
        $this->db_user = DB_USERNAME;
        $this->db_pass = DB_PASSWORD;
        $this->db_name = DB_NAME;

        $this->connection = new mysqli($this->db_host, $this->db_user, $this->db_pass, $this->db_name);
    }

    function connect($host, $user, $pass, $db_name) {

        $this->connection = new mysqli($host, $user, $pass, $db_name);

        if ( mysqli_connect_errno() ) {
            printf("Connection failed: %s", mysqli_connect_error());
            exit();
        }

        return $this->connection;

    }

    public function query($sql)
        {
        return $this->connection->query($sql);
        }

   }

//--------

require_once("connection.php");

class DB {

    public $con;

    function __construct()
    {
        $this->con = new connection();
    }

    function insertQuery($table, $data_array){ 

        $fields = " (";
        $values = " VALUES (";
        $x = 0;
        $y = 0;

        foreach ($data_array as $key => $value) {

            $fields.= $key;
            $values.= "'".addslashes($value)."'";

            if(sizeof($data_array)-1 != $x){
                $fields.= ",";
                $values.=",";
            }
            else{
                $fields.= ")";
                $values.=")";
            }
            $x++;
        }

        $sql = "insert into ".$table.$fields.$values;

        if (!$this->con->query($sql) === TRUE) {
            return "Error: " . $sql . "<br>" . $this->con->error;
        }
        else {
            $last_id = $this->con->insert_id;
            return $last_id;
        }
    }

}

Answer 1

首先在DB类构造函数中，您应该设置凭据属性（db_host, db_user, db_pass, db_name）并调用connection类的connection方法

class DB {

    public $con;

    function __construct()
    {
        $this->con = new connection();
        $this->con->db_host = HOST; // replace HOST
        $this->con->db_user = DB_USERNAME; // replace DB_USERNAME
        $this->con->db_pass = DB_PASSWORD; // replace DB_PASSWORD
        $this->con->db_name = DB_NAME; // replace DB_NAME
        $this->con->connection();
    }

    function insertQuery($table, $data_array){ 

        $fields = " (";
        $values = " VALUES (";
        $x = 0;
        $y = 0;

        foreach ($data_array as $key => $value) {

            $fields.= $key;
            $values.= "'".addslashes($value)."'";

            if(sizeof($data_array)-1 != $x){
                $fields.= ",";
                $values.=",";
            }
            else{
                $fields.= ")";
                $values.=")";
            }
            $x++;
        }

        $sql = "insert into ".$table.$fields.$values;

        if (!$this->con->query($sql) === TRUE) {
            return "Error: " . $sql . "<br>" . $this->con->error;
        }
        else {
            $last_id = $this->con->insert_id;
            return $last_id;
        }
    }
 }

无法访问php mysqli连接对象属性

1 个答案: