mysqli_insert_id错误

Question

Using the PHP manual I created following code:

$query = "INSERT INTO inserir(nome) VALUES ('Stefanato');";
$listar = new consultar();
$listar->executa($query);

echo "New record has id: " . mysqli_insert_id($listar->$query);

I also used this answer for class connections: Error mysqli_select_db

But I keep getting this error:

Warning: mysqli_insert_id () expects parameter exactly 1, 2 given in /home/controle/public_html/demo/teste.php on line 9

How do I fix that?

Answer 1

Here is the simple example for getting the id of last record created-

Code is taken from here

<?php
    $con=mysqli_connect("localhost","my_user","my_password","my_db");
    // Check connection
    if (mysqli_connect_errno()) {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    mysqli_query($con,"INSERT INTO Persons (FirstName,LastName,Age)
        VALUES ('Glenn','Quagmire',33)");

    // Print auto-generated id
    echo "New record has id: " . mysqli_insert_id($con);

    mysqli_close($con);
?> 

Edit: Here is working code

<?php
define("SERVIDOR_BD", "localhost");
define("USUARIO_BD", "usuario");
define("SENHA_BD", "senha");
define("BANCO_DE_DADOS", "dados");


class conecta {
    public $database_bancoDados = null;//1. New Added Field
    public $bancoDados = null;//2. New Added Field
    function conecta($servidor="", $bancoDeDados="", $usuario="", $senha=""){
        if (($servidor == "") && ($usuario == "") && ($senha == "") && ($bancoDeDados == "")){
            $this->bancoDados = mysqli_connect(SERVIDOR_BD, USUARIO_BD, SENHA_BD) or trigger_error(mysqli_error(),E_USER_ERROR);//3. Store in class variable
            $this->database_bancoDados = BANCO_DE_DADOS;//4. Store in class variable
        } else {
            $this->bancoDados = mysqli_connect($servidor, $usuario, $senha) or trigger_error(mysqli_error(),E_USER_ERROR);//5. Store in class variable
            $this->database_bancoDados = $bancoDeDados;//6. Store in class variable
        }
    }    
}

class consultar {

    var $bd;
    var $res;
    var $row;
    var $nrw;
    var $data;

    function executa($sql=""){
        if($sql==""){
            $this->res =  0; // Pointer result of the executed query
            $this->nrw =  0; // Line number the query returned, cruise control
            $this->row = -1; // Array of the current query line
        }
        // Connects to the database   
           $this->bd = new conecta();//7. Store in class variable
           $this->bd->conecta();//8. Store in class variable
           mysqli_select_db($this->bd->bancoDados, BANCO_DE_DADOS);//9. Change Here For parameter sequence
            $this->res = mysqli_query($this->bd->bancoDados, $sql); //10. Change here for parameter sequence

           $this->nrw = @mysqli_num_rows($this->res);

        $this->row = 0;
        if($this->nrw > 0)
            $this->dados();
    }

    function primeiro(){
        $this->row = 0;
        $this->dados();
    }

    function proximo(){
        $this->row = ($this->row<($this->nrw - 1)) ?
                        ++$this->row:($this->nrw - 1);
        $this->dados();
    }

    function anterior(){
        $this->row = ($this->row > 0) ? -- $this->row:0;
        $this->dados();
    }        

    function ultimo(){
        $this->row = $this->nrw-1;
        $this->dados();
    }

    function navega($linha){
        if($linha>=0 AND $linha<$this->nrw){
            $this->row = $linha;
            $this->dados();
        }
    }

    function dados(){
        mysqli_data_seek($this->res, $this->row);
        $this->data = mysqli_fetch_array($this->res);
    }
}

$query = "INSERT INTO inserir(uname) VALUES ('Stefanato');";
$listar = new consultar();
$listar->executa($query);

echo "New record has id: " . mysqli_insert_id($listar->bd->bancoDados);//11. Change Here for parameter