如何检查用户是否已登录 Symfony?

本文介绍了如何检查用户是否已登录 Symfony?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我收到了这张表格:

{{ form_start(form) }} <input type="submit" class="form_submit_button" name="register_interest" value="Add to favourites"/> {{ form_end(form) }}

我想检查用户是否登录，我这样做了:

I want to check if user is logged in and I did this :

{% if is_granted('IS_AUTHENTICATED_FULLY') %}

如果用户未登录，它会隐藏按钮.

And it hides the button if the user is not logged in.

即使我没有登录，我也希望该按钮显示，并在提交表单后检查用户是否登录.

I want that button to show even if I am not logged in and after the form submit to check if the user is logged in .

我该怎么做?

谢谢！

推荐答案

必须在您的 php 控制器中执行检查.此外，您还可以执行 javascript 检查，以防止在用户未登录时提交表单.

The check must be performed in your php controller. Additionally, you can also perform a javascript check which prevents submitting the form if user is not logged in.

要检查控制器中的用户登录:

To check for user login in your controller:

if ($this->isGranted('ROLE_USER') == false) { // show some error message, throw exception etc... }

另一种方法:

$this->denyAccessUnlessGranted('ROLE_USER');

要检查客户端 (javascript)，您可以执行以下操作:

To check on the client side (javascript), you could do something like this:

{% if not is_granted('IS_AUTHENTICATED_REMEMBERED') %} <script> // wrap into document.ready() if needed $('#your_submit_button_id').on('click', function(e) { e.preventDefault(); alert('You must be logged in to use this function'); }); </script> {% endif %}