如何使用symfony更新mysql列

Question

我在視圖文件html輸入表單中創建了一個表單，並將其命名為“ thenum”

{% extends 'base.html.twig' %}

{% block body %}
    <h1>User edit</h1>

    {{ form_start(edit_form) }}
        {{ form_widget(edit_form) }}
        <input type="text" name="thenum" value="0"><br>

        <input type="submit" value="Edit" />
    {{ form_end(edit_form) }}


{% endblock %}

並在我的默認控制器文件中

public function indexAction()
{

    $user = $this->container->get('security.context')->getToken()->getUser();
    $user->getId();
    $conn = $this->get('database_connection');

    $users = $conn->query("UPDATE user SET batman= 'thenum' WHERE username ='$user'");
    return $this->render('siteblogBundle:Default:index.html.twig', array('edit_form' => $users));
}

但是我運行代碼時收到此錯誤

類型錯誤：傳遞給Symfony \\ Component \\ Form \\ FormRenderer :: renderBlock（）的參數1必須是Symfony \\ Component \\ Form \\ FormView的實例，Doctrine \\ DBAL \\ Driver \\ PDOStatement的實例，在/ opt / lampp /中調用htdocs / x / chessMult / 1 / sym / thesym / app / cache / dev / twig / 96 / 96f6b6f8c21d5412246839a2d6e2eb66ac4141143dd67e2b91f285f2a31b60fc.php在第44行

Answer 1

您應該使用Symfony表單 。 表單中返回的數據可以刷新以更新數據庫記錄。 文檔中提供了一個示例。

$task = new Task();

$form = $this->createFormBuilder($task)
    ->add('task', TextType::class)
    ->add('dueDate', DateType::class)
    ->add('save', SubmitType::class, array('label' => 'Create Task'))
    ->getForm();

$form->handleRequest($request);

if ($form->isSubmitted() && $form->isValid()) {
    // $form->getData() holds the submitted values
    // but, the original `$task` variable has also been updated
    $task = $form->getData();

    // ... perform some action, such as saving the task to the database
    // for example, if Task is a Doctrine entity, save it!
    // $em = $this->getDoctrine()->getManager();
    // $em->persist($task);
    // $em->flush();

    return $this->redirectToRoute('task_success');
}

return $this->render('default/new.html.twig', array(
    'form' => $form->createView(),
));