PHP error when connecting to MySQL database

Question

I was trying to connect to my local database through the php code but I get this error:

Syntax error, unexpected '$result' (T_VARIABLE), expecting ',' or ')'

and I don't understand where's the problem. Here's my code:

    <?php
    $hostname = "localhost";
    $username = "root";
    $password = "";

    $databaseName = "newspage";

    $dbConnected = @mysqli_connect($hostname, $username, $password);

    $dbSelected = @mysqli_connect($databaseName, $dbConnected);

    $query = "INSERT INTO news(titolo, testo, data)VALUES('".$_POST["titolo"]."', '".$_POST["testo"]."', NOW())";
    $result = @mysqli_query($query);
    if(!$result){
    echo("Errore aggiunta news: " . mysqli_error(mysqli $result));
    exit();
    }

    else {
    mysqli_close(mysqli $dbConnected);
    echo('News caricata!<br><a href="add.php">Clicca qui</a> per aggiungere altre news.<br><a href="edit.php">Clicca qui</a> per apportare modifiche alle news.<br><a href="../index.php">Clicca qui</a> per tornare alla pagina principale.'); 
    }
?>

Learn to use prepared statements instead of concatenating variables, to protect against SQL injection. And don't use @ when you're still debugging. — Barmar
– Barmar, Commented Jul 12, 2018 at 22:14
Your second call to mysqli_connect() should be mysqli_select_db(), and the arguments are in the wrong order. — Barmar
– Barmar, Commented Jul 12, 2018 at 22:15
WARNING: Using the error-suppressing YOLO operator (@) obscures problems with your code and makes debugging issues like this a whole lot more complicated. That's a tool of last resort and should only be used in exceptional circumstances. You should display an error message for the user, log a problem, initiate some kind of retry, or all of these things in conjunction. — tadman
– tadman, Commented Jul 12, 2018 at 22:20
Note: The object-oriented interface to mysqli is significantly less verbose, making code easier to read and audit, and is not easily confused with the obsolete mysql_query interface. Before you get too invested in the procedural style it’s worth switching over. Example: $db = new mysqli(…) and $db->prepare("…") The procedural interface is an artifact from the PHP 4 era when mysqli API was introduced and should not be used in new code. — tadman
– tadman, Commented Jul 12, 2018 at 22:20
WARNING: When using mysqli you should be using parameterized queries and bind_param to add user data to your query. DO NOT use string interpolation or concatenation to accomplish this because you have created a severe SQL injection bug. NEVER put $_POST, $_GET or any user data directly into a query, it can be very harmful if someone seeks to exploit your mistake. — tadman
– tadman, Commented Jul 12, 2018 at 22:20

alanfcm · Accepted Answer · 2018-07-12 22:12:51Z

1

Your problem is in this line:

echo("Errore aggiunta news: " . mysqli_error(mysqli $result));

It should be:

echo("Errore aggiunta news: " . mysqli_error($result));

answered Jul 12, 2018 at 22:12

alanfcm

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2 Comments

Jasmine Over a year ago

If I edit it I get this: Warning: mysqli_error() expects parameter 1 to be mysqli, null given

alanfcm Over a year ago

That's because you are not running the query against any database.

Naveed Ramzan · Accepted Answer · 2018-07-12 22:09:56Z

0

You should pass dbname too in mysqli_connect

$dbConnected = @mysqli_connect($hostname, $username, $password, $databaseName);

and to execute query.

$result = $dbConnected->query("select * from tablename");

Further your query should be:

    $query = "INSERT INTO news(titolo, testo, data) 
    VALUES('".$_POST['titolo']."', '".$_POST['testo']."', NOW())";

answered Jul 12, 2018 at 22:09

Naveed Ramzan

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2 Comments

Barmar Over a year ago

Neither of these is related to the syntax error being reported.

tadman Over a year ago

That query is in urgent need of placeholder values.

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