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I’m trying to make a query directly in the PHP file where it creates the connection to the database. This query will create my table. The problem is that I am trying to reference the connection variable in mysqli_query() and it’s not being referenced (I believe this is the problem). I did it in this structure because I’m doing Object-Oriented!
The mistake you’re making is: Warning: mysqli_query() expects Parameter 1 to be mysqli, Boolean Given in...
Follows the code:
<?php
class db{
private $dbServer = "localhost";
private $dbUser = "root";
private $dbPassword = "";
private $dbName = "sql_queries";
private $conn;
public function __construct(){
$this->conn = mysqli_connect($this->dbServer, $this->dbUser, $this->dbPassword) or die("Não foi possível estabelecer a conexão com a base de dados! ".mysqli_error($this->conn));
$this->conn = mysqli_select_db($this->conn, $this->dbName) or die("Não foi possível se conectar com o banco de dados! ".mysqli_error($this->conn));
$this->setQuery();
}
public function getConn(){
return $this->conn;
}
public function setQuery(){
$sql = "CREATE TABLE IF NOT EXISTS sql_queries.aluno(
aluno_id INT NOT NULL AUTO_INCREMENT,
nome_aluno varchar(255),
dataNascimento_aluno date,
nome_pai varchar(255),
nome_mae varchar(255),
RG_aluno char(10),
cpf_aluno char(11),
telefone_aluno char(10),
celular_aluno char(13),
CONSTRAINT PK_aluno PRIMARY KEY (aluno_id)
);";
$result = mysqli_query($this->conn, $sql);
}
}
Perfect! After doing what you said it worked!!! I didn’t know that I didn’t need to reference twice, and that the function mysqli_select_db() is only for switching banks! I’ll keep the tip in mind!
– Wesley Redfield
William, actually, in this line of code $this->Conn = mysqli_select_db I’m overwriting the "Link" with another result... so it didn’t work right?
– Wesley Redfield
@Wesleyredfield that’s right, mysqli_select_db does not contain the link, it only returns true or false, to say whether you were able to change the database or not.
– Guilherme Nascimento