HTML / PHP表单不会将数据发送到MySQL数据库

Question

我是PHP的新手，所以这对我来说是新手。无论如何我正在创建一个网站，所以我可以在线访问我的receipes。

这是形式：

<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>

这是处理表单的代码：

<?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";

}

?>

会发生什么是如果我输入这样的东西它不起作用：

标题：AçordadeCamarao

类别：鱼

配料：虾仁800克; 4瓣大蒜; 1枝欧芹或香菜; 3个全蛋; 1.5分升橄榄油;每人1.5面包;盐;皮瑞 - 皮瑞·

准备：用盐和piri-piri烹饪虾并保留水。然后将面包放入虾的水中。用大蒜和香菜加热橄榄油，然后加入虾，最后加入面包。将所有东西混合在一起烤面包并获得一致性。最后，鸡蛋被收集起来，一切都被包裹起来。

注意：4人的收入

但如果我像这样输入它的工作原理：

标题：gfdsfdsa

类别：鱼

成分：hudsbfbdsf fdsfidsfidsfsd，fdsjifjdsifdis 0palpdsandnsaud jkdosakodsakodmnsa jidsjaidsa

准备工作：nfjdbshfbhdbjfdjs dsajijdisandiabuu fjndoisjfojidsanfds

注意：fbhdubsufbndsnfs

我的数据库表：

Nome    Tipo    Agrupamento (Collation) Atributos   Nulo    Omissao Extra   

1 ID bigint（50）否无AUTO_INCREMENT静音静音删除 2标题varchar（100）utf8_general_ci否无静音变量3 Varchar（10）utf8_general_ci无无静音静音 4成分varchar（1000）utf8_general_ci否无更改5准备varchar（1000）utf8_general_ci否无静音更改6注意varchar（1000）utf8_general_ci否无静音静音

对不起，如果这个帖子很长。任何想法如何解决它？

Answer 1

输入中可能存在一个特殊字符，导致查询中出现语法错误。

在将其替换为查询之前，您需要转义输入。在include ('connection.php');之后加上这个

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

或（更好）使用准备好的声明。用它来代替你对mysqli_query的调用：

$stmt = mysqli_prepare($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);