HTML/PHP Form doesn't sent data to MySQL Database

Question

I'm new to PHP so this is new to me. Anyway I'm creating a site so I can access my receipes online.

This is the form:

<form action="Form.php" method="post" class="basic-grey">
<h1><a href="index.html">Receita</a>
    <span>Aqui podes adicionar uma nova receita</span>
</h1>
<label>
    <span>Titulo :</span>
    <input id="titulo" type="text" size="20" maxlength="100" name="titulo" placeholder="Introduza o Titulo" />
</label>

<label>
    <span>Categoria :</span><select  maxlength="10" name="categoria">
    <option value="categoria">--- Seleccione aqui a Categoria ---</option>
    <option name=" " value="sopa">Sopa</option>
    <option name="entrada" value="entrada">Entrada</option>
    <option name="carne" value="carne">Carne</option>
    <option name="peixe" value="peixe">Peixe</option>
    <option name="salada" value="salada">Salada</option>
    <option name="sobremesa" value="sobremesa">Sobremesa</option>
    </select>

</label>

<label>
    <span>Ingredientes :</span>
    <textarea id="ingredientes" size="20" maxlength="1000" name="ingredientes" placeholder="Introduza os ingredientes"></textarea>
</label> 
<label>
    <span>Preparação :</span>
    <textarea id="preparacao" size="20" maxlength="1000" name="preparacao" placeholder="Introduza o modo de preparação"></textarea>
</label> 
<label>
    <span>Notas :</span>
    <textarea id="notas" size="20" maxlength="1000" name="notas" placeholder="Aqui pode adicionar uma nota"></textarea>
</label> 
<label>
    <span>&nbsp;</span> 
    <input type="submit" class="button" value="Enviar" /> 
</label>   

</form>

This is the code to handle the form:

<?php
// processing form values


if ($_SERVER['REQUEST_METHOD'] == 'POST'){

$titulo = $_POST['titulo'];
$categoria = $_POST['categoria'];
$ingredientes = $_POST['ingredientes'];
$preparacao = $_POST['preparacao'];
$notas = $_POST['notas'];


if(!empty($titulo) && !empty($categoria) && !empty($ingredientes) && !empty($preparacao) && !empty($notas)){

    include('connection.php');

    mysqli_query($dbc, "INSERT INTO receita(Titulo,Categoria,Ingredientes,Preparacao,Notas) VALUES ('$titulo','$categoria','$ingredientes','$preparacao','$notas')");
    $registered = mysqli_affected_rows($dbc);
    echo $registered." row is affected, everything worked fine!";
}else{
    echo "Please fill all values on the form";
}

}else{

echo "No form has been submitted";

}

?>

And what happens is that if I input something like this it doesn't work:

Titulo:Açorda de camarao

Categoria: Peixe

Ingredientes: 800 g de camarao; 4 dentes de alho; 1 ramo de salsa ou coentros; 3 ovos inteiros; 1.5 dl de Azeite; 1.5 pão por pessoa; sal; piri-piri

Preparação: Coze-se o camarão com sal e piri-piri e reserva-se a agua. De seguida demolha-se o pão na agua do camarão. Aquece-se o azeite com os alhos e os coentros e de seguida junta-se o camarão e por ultimo o pão. Mexe-se tudo para cozer o pão e ganhar consistencia. Por ultimo junta-se os ovos e envolve-se tudo.

Nota: Receita para 4 pessoas

But if I input like this it works:

Titulo:gfdsfdsa

Categoria: Peixe

Ingredientes: hudsbfbdsf fdsfidsfidsfsd, fdsjifjdsifdis 0palpdsandnsaud jkdosakodsakodmnsa jidsjaidsa

Preparação: nfjdbshfbhdbjfdjs dsajijdisandiabuu fjndoisjfojidsanfds

Nota: fbhdubsufbndsnfs

My database table:

Nome    Tipo    Agrupamento (Collation) Atributos   Nulo    Omissao Extra

1 ID bigint(50) Não None AUTO_INCREMENT Muda Muda Elimina
2 Titulo varchar(100) utf8_general_ci Não None Muda Muda Elimina 3 Categoria varchar(10) utf8_general_ci Não None Muda Muda
4 Ingredientes varchar(1000) utf8_general_ci Não None Muda 5 Preparacao varchar(1000) utf8_general_ci Não None Muda Muda 6 Notas varchar(1000) utf8_general_ci Não None Muda Muda

Sorry if this post its to long. Any ideas how to fix it?

Add error reporting to the top of your file(s) right after your opening PHP tag for example <?php error_reporting(E_ALL); ini_set('display_errors', 1); then the rest of your code, to see if it yields anything, as well as or die(mysqli_error($dbc)) to mysqli_query(). — Funk Forty Niner
– Funk Forty Niner, Commented May 6, 2015 at 15:53

Barmar · Accepted Answer · 2015-05-06 17:29:32Z

5

There's probably a special character in the input that's causing a syntax error in the query.

You need to either escape your input before substituting it into the query. Add this after include ('connection.php');

$titulo = mysqli_real_escape_string($dbc, $titulo);
$categoria = mysqli_real_escape_string($dbc, $categoria);
// and so on for all the other variables

or (better) use a prepared statement. Use this in place of your call to mysqli_query:

$stmt = mysqli_prepare($dbc, "INSERT INTO receita(Titulo,Categoria,Ingredientes,Preparacao,Notas) VALUES (?, ?, ?, ?, ?)");
mysqli_stmt_bind_param($stmt, "sssss", $titulo, $categoria, $ingredientes, $preparacao, $notas);
mysqli_stmt_execute($stmt);
$registered = mysqli_stmt_affected_rows($stmt);

edited May 6, 2015 at 17:29

answered May 6, 2015 at 15:59

Barmar

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4 Comments

Barmar Over a year ago

@BryanWay He's using mysqli in the question. I didn't change that.

Joao Mira Over a year ago

Sorry guys but i'm kinda lost. The code above from Barmar do i add it, or replace it? and where? sorry, noob here :p

Barmar Over a year ago

I've added more explanation, I hope it's enough. It should be clear if you understand what these things do.

Joao Mira Over a year ago

I did what you said and still doesn't work, don't know if i'm doing anything wrong...

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