symfony

问题 I am reading tutorial here https://symfony.com/doc/current/form/csrf_protection.html how to add csrf token. It says to use form_end() in the template. But this is not working, gives error: Type error: Too few arguments to function Symfony\Component\Form\FormRenderer::renderBlock(), 0 passed in E:\projektai\php projektai\htdocs\mokomieji\symfony_4_demo\var\cache\dev\twig\bb\bb2248f7be504240fcc2ab43dabf593090ebc4c897ce72b1a979082d62914b47.php on line 48 and at least 2 expected Here is answer

问题 so I've been following this Database and the Doctrine tutorial: https://symfony.com/doc/current/doctrine.html the only difference is that I added a created_ts field to it (among some other fields, but they work fine so no need to go into them). I used the make:entity command to generate my class and the method for setting my created_ts got generated like this: public function setCreatedTs(\DateTimeInterface $created_ts): self { $this->created_ts = $created_ts; return $this; } So in my /index

问题 so I've been following this Database and the Doctrine tutorial: https://symfony.com/doc/current/doctrine.html the only difference is that I added a created_ts field to it (among some other fields, but they work fine so no need to go into them). I used the make:entity command to generate my class and the method for setting my created_ts got generated like this: public function setCreatedTs(\DateTimeInterface $created_ts): self { $this->created_ts = $created_ts; return $this; } So in my /index

问题 Using LexikJWTAuthenticationBundle, FOSRest, FOSUser how do I get authenticated user profile by token. Is it possible? So let's say user is already authenticated via LexikJWT and I have an api endpoint like /api/profile where I send the token and I expect to get specified user data. I'm using for frontend ReactJS with Redux. 回答1: This is an example of how to get your user by a service when the user is already authenticated: class UserService { /** @var TokenStorageInterface */ private

问题 I try to clone the content element image-slider or image-gallery (the error will come at both) to extend them. First I register a new CmsElement like the original only changes the name from image-slider to image-slider-example import './component'; import './config'; import './preview'; Shopware.Service('cmsService').registerCmsElement({ name: 'image-slider-example', label: 'sw-cms.elements.imageSlider.label', component: 'sw-cms-el-image-slider', configComponent: 'sw-cms-el-config-image

问题 My task is (or was) simple: There is a variety but finite types of "Post", each with a vastly different set of properties while all sharing a couple core properties (like: Title, Content, Published, etc) What I set out to do, was create a base Post entity with all the shared properties (as mentioned above), and have this entity as sort of a "Governor" entity, but more on that in a second. All the different Category entities would extend off of this base Post entity. I did it this way mostly

问题 My task is (or was) simple: There is a variety but finite types of "Post", each with a vastly different set of properties while all sharing a couple core properties (like: Title, Content, Published, etc) What I set out to do, was create a base Post entity with all the shared properties (as mentioned above), and have this entity as sort of a "Governor" entity, but more on that in a second. All the different Category entities would extend off of this base Post entity. I did it this way mostly

问题 I have provided a pre build project on symfony in which the logout session redirects to the login screen, but now i want that page to redirect on the home page instead. What i have found in the coding files is this: in the base twig file: <a href="{{path('log_out')}}"><i class="icon-key"></i> Log Out</a> in routing.yml #Route for logout page. log_out: pattern: /bid/logout Do anyone know how to change this redirection please hep me out, i am a total newbie to symfony Thanks 回答1: Normally it is

问题 I have provided a pre build project on symfony in which the logout session redirects to the login screen, but now i want that page to redirect on the home page instead. What i have found in the coding files is this: in the base twig file: <a href="{{path('log_out')}}"><i class="icon-key"></i> Log Out</a> in routing.yml #Route for logout page. log_out: pattern: /bid/logout Do anyone know how to change this redirection please hep me out, i am a total newbie to symfony Thanks 回答1: Normally it is

问题 I am implementing an alphabetical search. We display a table of Names. I want to highlight only those alphabets, which have names that begin with the corresponding alphabet. I am stumped with a simple problem. How to read the first character in the string user.name within twig. I have tried several strategies, including the [0] operation but it throws an exception. Here is the code {% for i in ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y'