title: How do I pass Parameters in Dynamic Links?
body: <p>How to receive parameters from firebase dynamic links in flutter?</p>
  
  <p>I created a short url:</p>
  
  <blockquote>
    <p><a href="https://subdomain.example.com/product" rel="noreferrer">https://subdomain.example.com/product</a></p>
  </blockquote>
  
  <p>which points to</p>
  
  <blockquote>
    <p><a href="https://example.com/view-product" rel="noreferrer">https://example.com/view-product</a></p>
  </blockquote>
  
  <p>But I want to add a url query parameter like:</p>
  
  <p><a href="https://example.com/view-product?id=56" rel="noreferrer">https://example.com/view-product?id=56</a></p>
  
  <p>Note that "56" is variable and changes dynamically inside app flow. I am unable to receive this "id" parameter.</p>
  
  <p>On browser I tried entering <a href="https://subdomain.example.com/product?id=56" rel="noreferrer">https://subdomain.example.com/product?id=56</a></p>
  
  <p>I received the link: <a href="https://example.com/view-product" rel="noreferrer">https://example.com/view-product</a></p>
  
  <pre><code>     FirebaseDynamicLinks.instance.onLink(
          onSuccess: (PendingDynamicLinkData dynamicLink) async {
  
        final Uri deepLink = dynamicLink?.link;
  
        showModalBottomSheet(context: context, builder: (context){
          return Container(
            height: 100.0,
            child: Text(deepLink.toString()),
          );
        });
        if (deepLink != null) {
          debugPrint("Link found on line: "+deepLink.queryParameters.toString());
  
        }
      }, onError: (OnLinkErrorException e) async {
        print('onLinkError');
        print(e.message);
      });
  </code></pre>