samedi 11 février 2017

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, connector in a function

I'm trying to output something like, if user exists in the database then they can't register but I'm getting an error, it returns false instead of an actual result, I don't understand why.

Here's my connector function

public function db_connect(){
        $db_connect = mysqli_connect(
                $this->db_host,
                $this->db_user,
                $this->db_pass,
                $this->db_name
                );
            if (mysqli_connect_errno($db_connect)) {
                echo "<font color='red'>"."Failed to connect to MySQL: " . mysqli_connect_error() ."</font></br>";
            }
            return $db_connect;
        }

Here's where the validation goes.

public function testtable_add($Testtable1, $Testtable2, $Testtable3, $Flag){
            #$Id = $this->id;
            $this->testtable1 = $Testtable1;
            $this->testtable2 = $Testtable2;
            $this->testtable3 = $Testtable3;
            $this->flag = $Flag;

            $connector = new connector();


            $query = "SELECT * FROM testtable WHERE testtable1 = $Testtable1";
            $result = mysqli_query($connector->db_connect(), $query);

            while($row = mysqli_fetch_assoc($result)){
                if($result->num_rows){
                    echo "Data exists!";
                } else {

                    $query = "INSERT INTO test_table (testtable1, testtable2, testtable3, flag) VALUES ('$this->testtable1','$this->testtable2','$this->testtable3', '$this->flag')";

                }
            }

    }

The error occur upon insertion of data.




Aucun commentaire:

Enregistrer un commentaire