mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement

Question

I am They try to use mysqli :: bind_param unsuccessfully . In the code , below , the function login ( ) is called with a username and password on the page login.php . Despite all efforts and guides and forums have read , continues to return the same error .

I tried both with

bind_param ( 's' , $ variable )

that with

bind_param ( 1 , $ variable )

that with

bind_param ( 'ss' , $ variable1 , $ variable2 )

and i tried query without ''

"SELECT id,org_id,org_group_id,people_id FROM users WHERE username = ? AND password = ?"

Where am I wrong ?

public function login($_username, $_password) {
    $this->sessionOpen ();

     if ($_username == "") {
        $this->log->error ( "Username vuoto" );
        throw new AuthLoginFailed ();
    }
    if ($_password == "") {
        $this->log->error ( "Password vuota" );
        throw new AuthLoginFailed ();
    }

    $db = new mysqli ( $this->sql ['server'], $this->sql ['username'], $this->sql ['password'], $this->sql ['database'] );
    if (mysqli_connect_errno ()) {
        $this->log->error ( "Errore di connessione a mysql: " . mysqli_error ( $db ) );
        throw new MysqliConnectionError ( "Mysqli error: " . mysqli_error ( $db ) );
    }

    $stmt = $db->prepare ( "SELECT id,org_id,org_group_id,people_id FROM users WHERE 'username' = ? AND 'password' = ?" );
    if (! $stmt) {
        $this->log->error ( "Mysqli prepare error: " . mysqli_error ( $db ) );
        throw new MysqliPrepareException ( "Mysqli error: " . mysqli_error ( $db ) );
    }
    echo md5 ( $_username ) . "---" . md5 ( $_password );
    //on page username and password is showed at this point
    $user=trim(md5 ( $_username ));
    $pass=trim(md5 ( $_password ));
    $stmt->bind_param ( 1,  $user);
    $stmt->bind_param ( 2,  $pass);
    /* Execute it */
    $stmt->execute ();
    if (! $stmt) {
        $this->log->error ( "Mysqli prepare error: " . mysqli_error ( $db ) );
        throw new MysqliExecuteException ( "Mysqli error: " . mysqli_error ( $db ) );
    }


    $stmt->fetch($rst);

    echo "results: " . $rst->num_rows; //output of this: results:

    if ($rst->num_rows == 0) {
        throw new AuthLoginFailed ();
    }



    /* Close statement */
    $stmt->close ();

    /* Close connection */
    $db->close ();
}

Error in the log of apache is

[Sat Oct 24 08:52:04 2015] [error] [client 192.168.253.6] PHP Warning:  mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement in /www/gexy/XXXX/html/lib/Auth.php on line 77, referer: https://gexy.it/login.php
[Sat Oct 24 08:52:04 2015] [error] [client 192.168.253.6] PHP Warning:  mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement in /www/gexy/XXXX/html/lib/Auth.php on line 78, referer: https://gexy.it/login.php

Many thanks to all

Jitendra Kumar. Balla · Accepted Answer · 2015-10-24 08:06:44Z

1

Modification is:

$stmt->bind_param ( "ss", $user, $pass); because 1 data type is not defind in bind_param (). bind_param() will take two arguments 1st one is types (i, d, s, b) corresponding datatype in your query(?) and 2nd arg are values.

Suggestion's are:

Don't compare with ==, for empty string because if user enter's 3 white spaces it will not equal. use empty() for checking empty string or not.
Don't call unnecessary methods, it does not have any meaning, for eg: in your code your calling trim() after md5(). md5() will not return any white space character. So calling trim(md5($username)) is meaning less.

Try to replace your code with my code hope your problem is solved.

public function login($_username, $_password) {
$this->sessionOpen ();

 if (empty($_username)) {
    $this->log->error ( "Username vuoto" );
    throw new AuthLoginFailed ();
}
if (empty($_password)) {
    $this->log->error ( "Password vuota" );
    throw new AuthLoginFailed ();
}

$db = new mysqli ( $this->sql ['server'], $this->sql ['username'], $this->sql ['password'], $this->sql ['database'] );
if (mysqli_connect_errno ()) {
    $this->log->error ( "Errore di connessione a mysql: " . mysqli_error ( $db ) );
    throw new MysqliConnectionError ( "Mysqli error: " . mysqli_error ( $db ) );
}

$stmt = $db->prepare ( "SELECT id,org_id,org_group_id,people_id FROM users WHERE 'username' = ? AND 'password' = ?" );
if (! $stmt) {
    $this->log->error ( "Mysqli prepare error: " . mysqli_error ( $db ) );
    throw new MysqliPrepareException ( "Mysqli error: " . mysqli_error ( $db ) );
}
echo md5 ( $_username ) . "---" . md5 ( $_password );
//on page username and password is showed at this point
$user=md5 ( $_username );
$pass=md5 ( $_password );
$stmt->bind_param ( "ss",  $user,$pass);
/* Execute it */
$stmt->execute ();
if (! $stmt) {
    $this->log->error ( "Mysqli prepare error: " . mysqli_error ( $db ) );
    throw new MysqliExecuteException ( "Mysqli error: " . mysqli_error ( $db ) );
}


$stmt->fetch($rst);

echo "results: " . $rst->num_rows; //output of this: results:

if ($rst->num_rows == 0) {
    throw new AuthLoginFailed ();
}



/* Close statement */
$stmt->close ();

/* Close connection */
$db->close ();
}

Let me know once your problem is solved.

edited Oct 24, 2015 at 8:06

answered Oct 24, 2015 at 7:57

Jitendra Kumar. Balla

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2 Comments

Eugenio Liguori Over a year ago

It works , and thank you for the other tips . How I see my mistake is quoting , I used ' ss ' instead of " ss ".

Eugenio Liguori Over a year ago

I also found another error bind_result , in the query I requested more than one column and I should specify more than one variable , corresponding to the required columns in bind_result . Thanks again

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