这个 PDO 脚本有什么问题

Question

这里我正在编写一个脚本，它将使用简单的 html form.But 将三个不同的值发送到数据库，每当我提交表单时都没有插入任何值，并且出现以下错误：

这可能是什么原因error.How我可以解决这个问题吗？

$server='localhost';
$user='root';
$password='';
$dbname='fruits';
if(isset($_POST['name']) &&isset($_POST['colour']) && isset($_POST['calories'])){
   if(!empty($_POST['name']) && !empty($_POST['colour']) && !empty($_POST['calories'])){
$name=$_POST['name'];
$colour=$_POST['colour'];
$calories=$_POST['calories'];
try{
$conn=new PDO('mysql:host=$server,dbname=$dbname,$user,$password');
$conn->setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);
$stmt=$conn->prepare("INSERT INTO favfruit(name,colour,calories) 
                      VALUES(:name,:colour,:calories)");

$stmt->execute(array(':name'=>$name,':colour'=>$colour,':calories'=>$calories));
echo 'it was successfully entered to database';
}catch(PDOException $e){
    echo 'error : '.$e->getMessage();
}
}
}

HTML 形式 :

<form action='<?php echo $_SERVER['PHP_SELF']; ?>' method='post'>
    fruit name :<input type='text' placeholder='fruit name' name='name'></br>
    colur      :<input type='text' placeholder='fruit colour' name='colour'></br>
    calories   :<input type='text' placeholder='calories' name='calories'></br>
    <input type='submit' value='SUBMIT FORM'>
</form>

Answer 1

require_once($_SERVER['DOCUMENT_ROOT'].'/settings.php');
class db extends pdo{
    //variables
    public $db = '';
    public $config;
    public $settings = array(
        'host'      => SERVER,
        'database'  => DB,
        'username'  => USER,
        'password'  => PASS,
    );

    public function __construct(){

        $this->db = new PDO(
            "mysql:host={$this->settings['host']};" .
            "dbname={$this->settings['database']};" .
            "charset=utf8",
            "{$this->settings['username']}",
            "{$this->settings['password']}"
        );

        $this->db->setAttribute(PDO::ATTR_DEFAULT_FETCH_MODE, PDO::FETCH_ASSOC);
    }

在 settings.php 文件中定义数据库连接的常量然后创建一个新函数。

 public function InsertFruit() {
 if(isset($_POST['name']) &&isset($_POST['colour']) && isset($_POST['calories'])){
       if(!empty($_POST['name']) && !empty($_POST['colour']) && !empty($_POST['calories'])){
    $name=$_POST['name'];
    $colour=$_POST['colour'];
    $calories=$_POST['calories'];
    try{
    $query = <<<SQL
    INSERT INTO favfruit(name,colour,calories)
    VALUES(:name,:colour,:calories)
SQL;
    $resource = $this->db->prepare($query);
    $resource->execute(array(
    'name'  => $_POST['name'],
    'colour' => $_POST['colour'],
    'calories' => $_POST['calories'],
));
    echo $_POST['name'].' entered into database';
    $stmt=$conn->prepare("INSERT INTO favfruit(name,colour,calories) 
                          VALUES(:name,:colour,:calories)");

    }catch(PDOException $e){
        echo 'error : '.$e->getMessage();
    }
    }
    }
    }

我肯定知道只要您的网络服务器已启动并且运行这将适用于您的连接，并且它将简化您以后需要使用的任何其他连接。至于你的实际形式，只需使用案例流程来确定什么功能是运行.

Answer 2

你的问题只是一个错字。替换

$conn=new PDO('mysql:host=$server,dbname=$dbname,$user,$password');

来自

$conn=new PDO("mysql:host=$server;dbname=$dbname",$user,$password);

这个 PDO 脚本有什么问题

what's wrong in this PDO script

php

mysql

forms

pdo