PHP Error - Call to a member function execute() on a non-object

Question

I wrote thte following piece of code:

<?php

$username = $_POST['user'];
$password = $_POST['pass'];

$db = new PDO ('mysql:host=localhost;dbname=ozdatabase;charset=utf8', 'root', '');
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
$db->setAttribute(PDO::ATTR_ERRMODE, 'ERRMODE_EXCEPTION');

$stmt = $db->prepare("SELECT id, users FROM ozusers WHERE username=? AND password=?");
$stmt->execute(array($username, $password));
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);

$id = $rows['id'];
$user = $rows['users'];

if ($id) {
    print "Logged";
}

else {
    print "not good";
}

?>

This is the HTML Form:

<form id='login' action='login.php' method='post' accept-charset='UTF-8'>
<fieldset >
<LEGEND>COMMUNICATION</LEGEND>
<input type='hidden' name='submitted' id='submitted' value='1' />

<label for='username' >UserName*:</label>
<input type='text' name='user' id='username' maxlength="50" />

<label for='password' >Password*:</label>
<input type='password' name='pass' id='password' maxlength="50">

<input type='submit' name='Submit' value='Submit' />

</fieldset>
</form>

I get an error when trying to login in the page and it's written: "Fatal error: Call to a member function execute() on a non-object in.. On line 15"

Why is this happening? I followed best practice guide and it showed to use the "execute()" function exactly like that..

Thanks

It is possible that $db->prepare failed and $stmt value is false. — Mauro Tamm
– Mauro Tamm, Commented Nov 17, 2015 at 12:59
betting your form's failing you. the one you didn't include. — Funk Forty Niner
– Funk Forty Niner, Commented Nov 17, 2015 at 13:00
It is possible that the PDO connection itself is invalid. New PDO object is created and prepare can be called - but prepare fails and returns false. FALSE is not a object so you cannot call false->execute. Doesn't your database require password and username besides root and ''? — Mauro Tamm
– Mauro Tamm, Commented Nov 17, 2015 at 13:10

Saty · Accepted Answer · 2015-11-17 13:24:40Z

3

ERRMODE_EXCEPTION is constant wrap off quotes from it

$db->setAttribute(PDO::ATTR_ERRMODE, ERRMODE_EXCEPTION);//Your code fails at this line

answered Nov 17, 2015 at 13:24

Saty

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3 Comments

Funk Forty Niner Over a year ago

Yeah I think this makes sense. Let's see what the OP has to say about it. Good catch btw.

Saty Over a year ago

Your answer is also helpful specially for error reporting and for password protection

Funk Forty Niner Over a year ago

Thanks, but I risk being downvoted for it. I'll pass, cheers

Mauro Tamm · Accepted Answer · 2015-11-17 13:14:46Z

0

Wrap your database init in try catch to capture any connection failures.

try {
    $db = new PDO($dsn, $user, $password);
} catch (PDOException $e) {
    echo 'Connection failed: ' . $e->getMessage();
}

Currently the db->prepare() is failing, returning false and therefor not allowing you to call execute on a non object.

answered Nov 17, 2015 at 13:14

Mauro Tamm

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Comments

Rehmat · Accepted Answer · 2015-11-17 13:35:56Z

0

<?php

    //error_reporting(0);

    $username = $_POST['user'];
    $password = $_POST['pass'];

    // Connecting, selecting database
    $db = new PDO ('mysql:dbhost=localhost;dbname=ozdatabase;charset=utf8', 'root', '');
    $db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
    $db->setAttribute(PDO::ATTR_ERRMODE, ERRMODE_EXCEPTION);

    //first
    $stmt = $db->prepare("SELECT id, users FROM ozusers WHERE username = :username AND password = :password");
    $stmt->execute(array(':username'=>$username, ':password' => $password));
    $rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
    foreach($rows AS $r) {
        $id = $r['id'];
        $user = $r['users'];
    }

    if ($id) {
        print "Logged";
    }

    else {
        print "not good";
    }

?>

edited Nov 17, 2015 at 13:35

answered Nov 17, 2015 at 12:57

Rehmat

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5 Comments

pwnphpown Over a year ago

Hey thanks for the answer, u have changed the "?" that I inserted to the query. From my understanding "?" is treated as the placeholder and should give me the ability to execute() by passing parameters.

Funk Forty Niner Over a year ago

@pwnphpown don't ignore comments under your question.

Rehmat Over a year ago

@pwnphpown Yeah ? is used as placeholder. Named placeholders can be used too like I did in above change.

Rehmat Over a year ago

@pwnphpown: Check for comments made on your question. You will get to know why your query wasn't working. Fred has mentioned something there

Rehmat Over a year ago

@pwnphpown check the updated answer. Confirm that your database name is correct and database / table / rows exist. Double check the provided details

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