Function not working

Question

The following function is not working and i cannot see why.

function nuevoContacto($_POST) {
    try {
        include('func/usarBases.php');
        $mensaje="INSERT INTO `t_contactos`(`id_c`, `nombre`, `telefono`, `telefono2`, `corto`, `celular1`, `celular2`, `email`, `puesto`, `id_a`) VALUES (NULL,'$_POST[nombre]','$_POST[tel1]','$_POST[tel2]','$_POST[corto]','$_POST[cel1]','$_POST[cel2]','$_POST[email]','$_POST[puesto]','$_POST[id_a]')";
        $hacerConsulta = $base->prepare($mensaje);
        $hacerConsulta->execute();
    }
    catch( PDOException $e) {
        echo "<p>Error Connection: " .$e->getMessage()."</p>";
    }   
    $hacerConsulta=null;
}

Once it is called the code breaks and nothing further is executed. but when you use it inside the main code it works

Sorry i reedited the source and then is still not working, in the include usarBases.php is the conector pdo called $base

You'll need to define `$base` inside your function. As well as `$instruccion`. — andrewsi, Aug 28 '13 at 14:48
For future reference, you should also give information about what error you get rather than just saying "it breaks" — Anigel, Aug 28 '13 at 14:49
`$_POST` is a superglobal. you should **NOT** not be naming a function parameter with that name. — Marc B, Aug 28 '13 at 14:50
Consider the following http://stackoverflow.com/a/6558736/1415724 and http://stackoverflow.com/questions/1660912/passing-post-array-to-php-function which outlines what `Marc B` mentioned above. — Funk Forty Niner, Aug 28 '13 at 14:51
Sorry for this code, but i'm self learning to do. this is the first time i ask for help in a year of learning — Cesar Peña, Aug 28 '13 at 15:20

score 2 · Answer 1 · answered Aug 28 '13 at 14:53

What it have to be

function nuevoContacto($base)
{
    $sql  = "INSERT INTO  t_contactos VALUES (NULL,?,?,?,?,?,?,?,?,?)";
    $data = array(
        $_POST['nombre'],
        $_POST['tel1'],
        $_POST['tel2'],
        $_POST['corto'],
        $_POST['cel1'],
        $_POST['cel2'],
        $_POST['email'],
        $_POST['puesto'],
        $_POST['id_a']
    );
    $stmt = $base->prepare($sql);
    $stmt->execute($data);
}

have to be called with $base as a parameter instead of $_POST

MisterBla · Accepted Answer · 2013-08-28T14:54:08.563

0

You're lacking a database connection in your function. Add the following to the very beginning of your function:

global $base;

When you add global $base to your function you'll be able to use it within your function without having to re-write the whole thing.

Unrelated note, but worth mentioning.

You are open to SQL injections and you're not using prepared statements as you should. You should be using placeholders and binding them later instead of passing they directly into your query.

And a tip for next time:
State in your question what isn't working. What your expectation is and what actually happens.

edited Aug 28 '13 at 14:54

answered Aug 28 '13 at 14:48

MisterBla

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Whoever downvoted my answer, explain why if you would. – MisterBla Aug 28 '13 at 14:55
1

It wasn't me, but I'm guessing it was because you used `global`, instead of passing it through the function. – Ben Fortune Aug 28 '13 at 14:56
@BenFortune Is using `global` considered back practice or something? – MisterBla Aug 28 '13 at 14:56
From what I've seen, SO hates `global`. :P – Ben Fortune Aug 28 '13 at 14:57
@BenFortune [This answer](http://stackoverflow.com/questions/5166087/php-global-in-functions#5166527) makes a good point. I can now see why my answer got downvoted. – MisterBla Aug 28 '13 at 15:00
Sorry for this code, but i'm self learning to do. this is the first time i ask for help in a year of learning – Cesar Peña Aug 28 '13 at 15:21
@CesarPeña I'd recommend you to take Your Common Sense's approach to be honest. – MisterBla Aug 28 '13 at 15:52

Function not working

2 Answers2