Object of class mysqli could not be converted to string in line 5

Question

I tried to look for questions with the same issue but I couldn't get their solutions to fit my code.

I keep getting the error:

Object of class mysqli could not be converted to string in line 5

Code:

<?php

    function aggiornamento($utente) {
        global $conn;   
        global $_CONFIG;

        $selezione = mysqli_query ($conn, "SELECT * FROM ".$_CONFIG['db_account'].".account WHERE login = '".$utente."' LIMIT 1");

        while ($account = mysqli_fetch_array($selezione)) {
            $_SESSION['IShop_Login']= $account['login'];
            $_SESSION['IShop_DR'] = $account['dr'];
            $_SESSION['IShop_DB'] = $account['db'];
            $_SESSION['IShop_AID'] =   $account['id'];
            $_SESSION['IShop_Admin'] = $account['Admin_IShop'];
        }
    }

?>

I tried a few solutions from already asked questions but to no avail. So I'm kindly asking you to correct my code for me and maybe a little explanation of what's going on so I'd learn.

My $conn:

$conn = mysql_connect($_CONFIG['host'], $_CONFIG['user'], $_CONFIG['pass']);

Whereas $_CONFIG['host'], $_CONFIG['user'], $_CONFIG['pass'] defined as:

$_CONFIG['host'] = "SERVER IP";
$_CONFIG['user'] = "root";
$_CONFIG['pass'] = "PW";

And most importantly, $_CONFIG['db_account'] = "account";. But I strongly believe the problem isn't with my $conn, I could be wrong tho.

Answer 1

There are some problems you should take care about

1. Try to avoid using global variables inside functions. It may be dangerous in any application because you might change them without wanting it.
2. Take care of SQL injection. You may consider starting to use prepared statements in order to avoid it.
3. You are limiting the results to 1 but still use a loop to get the results
4. You are using mysqli functions but the connection is done using mysql_connect
5. You didn't select any database to work with, when created the connection

The connection should be done using mysqli_connect:

$mysql = mysqli_connect($host, $user, $password, $databaseName);

Answer 2

I don't know if this will help?

General `mysqli` connection:
----------------------------

   $host    =   'localhost';
   $uname   =   'root'; 
   $pwd     =   'xxx'; 
   $db      =   'xxx';

   $conn=new mysqli( $host, $uname, $pwd, $db );


function aggiornamento( $utente=false ){
    global $conn;
    global $_CONFIG;

    if( !isset( $utente ) or empty( $utente ) ) return false;

    $sql="select `login`,`dr`,`db`,`id`,`admin_ishop` from `{$_CONFIG['db_account']}`.`account` where `login`=? limit 1;";

    $stmt=$conn->prepare( $sql );
    $stmt->bind_param( 's', $param );
    $param=$utente;

    $result=$stmt->execute();
    $stmt->bind_result( $login, $dr, $db, $id, $admin_ishop );
    $stmt->fetch();

    if( $result ){

        $_SESSION['IShop_Login']    = $login;
        $_SESSION['IShop_DR']       = $dr;
        $_SESSION['IShop_DB']       = $db;
        $_SESSION['IShop_AID']      = $id;
        $_SESSION['IShop_Admin']    = $admin_ishop;

        $stmt->close();
        return true;
    }
    $conn->close();
    return false;
}

Test

<?php

    include __DIR__.'\conn.php';
    /*
        Where conn.php is simply:
        -------------------------

        <?php
        $host   =   'localhost';
        $uname  =   'root'; 
        $pwd    =   'xxx'; 
        $db     =   'so_experiments';

        $conn   =   new mysqli( $host, $uname, $pwd, $db );
        ?>
    */

    /* Defined so as not to break sql stmt */
    $_CONFIG['db_account']='so_experiments';


    function aggiornamento( $utente=false ){
        global $conn;
        global $_CONFIG;

        if( !isset( $utente ) or empty( $utente ) ) return false;

        $sql="select `login`,`dr`,`db`,`id`,`admin_ishop` from `{$_CONFIG['db_account']}`.`account` where `login`=? limit 1;";

        $stmt=$conn->prepare( $sql );
        $stmt->bind_param( 's', $param );
        $param=$utente;

        $result=$stmt->execute();
        $stmt->bind_result( $login, $dr, $db, $id, $admin_ishop );
        $stmt->fetch();

        if( $result ){

            $_SESSION['IShop_Login']    = $login;
            $_SESSION['IShop_DR']       = $dr;
            $_SESSION['IShop_DB']       = $db;
            $_SESSION['IShop_AID']      = $id;
            $_SESSION['IShop_Admin']    = $admin_ishop;

            /* only for demo debug */
            echo 'login:', $login,', db:', $db,', dr:', $dr,', id:', $id,', admin_ishop:', $admin_ishop;

            $stmt->close();
            return true;
        }
        $conn->close();
        return false;
    }

    /* call the function with a value for `$utente` = 'default' */
    call_user_func( 'aggiornamento', 'default' );

    /* 
        outputs: 
        login:default, db:store, dr:43, id:1, admin_ishop:1
    */
?>

Answer 3

You are using mysqli_query but mysql_connect, so use mysqli_connect() and it works

Answer 4

You are mixing mysql and mysqli-functions, and this is known to fail.

Try replace your mysql_connect with mysqli_connect, but be aware it might require other/additional parameters.

Also, you should reconsider your usage of global variables.

Answer 5

Try this

<?php
function aggiornamento($utente){
global $conn;   
global $_CONFIG;
       $selezione = mysqli_query ($conn, "SELECT * FROM ".$_CONFIG['db_account'].".account WHERE login = '".$utente."' LIMIT 1");
       while ($row = $selezione->fetch_assoc())
       {
$_SESSION['IShop_Login']= $row['login'];
$_SESSION['IShop_DR'] = $row['dr'];
$_SESSION['IShop_DB'] = $row['db'];
$_SESSION['IShop_AID'] =   $row['id'];
$_SESSION['IShop_Admin'] = $row['Admin_IShop'];
       }
}
?>