title: LazyVGrid onTapGesture navigate to next screen swiftUI
body: <p>I am quite new to swiftUI. I have created a grid view on tapping on which I want to go to next screen. But somehow I am not able to manage to push to next screen. I am doing like this:</p>
  <pre><code>    var body: some View {
      NavigationView {
          ScrollView {
              LazyVGrid(columns: gridItems, spacing: 16) {
                  ForEach(viewModel.pokemon) { pokemon in
                      PokemonCell(pokemon: pokemon, viewModel: viewModel)
                          .onTapGesture {
                              NavigationLink(destination: PokemonDetailView(pokemon: pokemon)) {
                                  Text(pokemon.name)
                              }
                          }
                  }
              }
          }
          
          .navigationTitle(&quot;Pokedex&quot;)
      }
      
  }
  </code></pre>
  <p>Upon doing like this, I am getting a warning stating</p>
  <blockquote>
  <p>Result of 'NavigationLink&lt;Label, Destination&gt;' initializer is unused</p>
  </blockquote>
  <p>Can someone please guide me, how to do this?</p>