Symfony - generate url with parameter in controller

Question

I want to generate a Url directly in my controller  I want to user a url defined in my routing yml file that needs a parameter   I ve found that code in the Cookbook  Routage section      params    router- gt match   blog my-blog-post       array  slug    gt   my-blog-post     controller    gt   AcmeBlogBundle Blog show     uri    router- gt generate  blog show   array  slug    gt   my-blog-post         blog my-blog-post   But I don t understand to what is refering the  router  Obviously  it doesn t work  Is there a simple way to generate a routing url with a paramter in a controller

User · Answer

make sure your controller extends Symfony Bundle FrameworkBundle Controller Controller   you should also check app console debug router in terminal to see what name symfony has named the route  in my case it used a minus instead of an underscore  i e blog-show   uri    this- gt generateUrl  blog-show     slug    gt   my-blog-post

User · Answer

Get the router from the container    router    this- gt get  router      Then use the router to generate the Url   uri    router- gt generate  blog show   array  slug    gt   my-blog-post

User · Answer

It s pretty simple    public function myAction          url    this- gt generateUrl  blog show   array  slug    gt   my-blog-post         Inside an action   this- generateUrl is an alias that will use the router to get the wanted route  also you could do this that is the same     this- gt get  router  - gt generate  blog show   array  slug    gt   my-blog-post

User · Answer

If you want absolute urls  you have the third parameter    product url    this- gt generateUrl  product detail        array           slug    gt   slug             UrlGeneratorInterface  ABSOLUTE URL      Remember to include UrlGeneratorInterface   use Symfony Component Routing Generator UrlGeneratorInterface

[symfony] Symfony - generate url with parameter in controller

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