title: Handle same request multiple times
body: <p>I'm building a silly music quiz game for learning. I need to populate my view with related music from <a href="https://developers.deezer.com/api" rel="nofollow noreferrer">deezer api</a>. </p>
  
  <p>What I need:</p>
  
  <ol>
  <li>Get a random genre</li>
  <li>Get 5 artist from this genre (id + name)</li>
  <li>Get 1 music from each artist (name + link preview)</li>
  </ol>
  
  <p>So, <a href="https://developer.mozilla.org/pt-BR/docs/Web/API/Request" rel="nofollow noreferrer">I found my way</a> until step 3</p>
  
  <p>But I can't find out how to properly send the same request 4 times (for each artist), and my reasearch gave me nothing so far</p>
  
  <pre><code>function deezer() {
  
      const reqGenero = new Request('https://api.deezer.com/genre');
  
      fetch(reqGenero)
          .then(response =&gt; {
              if (response.status === 200) {
                  return response.json();
              } else {
                  throw new Error('Erro ao pegar gêneros');
              }
          })
          .then(generos =&gt; {
              /* pega genero aleatorio */
              var generoId = generos.data[Math.floor(Math.random() * 10 + 1)].id;
              //console.log('\ngenero... ' + generoId);
              return fetch('https://api.deezer.com/genre/' + generoId + '/artists')
          })
          .then(response =&gt; {
              if (response.status === 200) {
                  return response.json();
              } else {
                  throw new Error('Erro ao pegar artistas');
              }
          })
          .then(artistas =&gt; {
              /* 1 música de 4 artistas */
              var artistasIds = [];
              for(var i = 0; i &lt;= 4; i++) {   
                  artistasIds.push(artistas.data[i].id);
                  console.log('\nId: ' + artistasIds[i]);
  
                  // CAN I SEND THIS REQUEST 4 TIMES?
                  return fetch('https://api.deezer.com/artist/' + ids + '/top'); 
              }
          })         
          .catch(error =&gt; {
              console.error(error);
          });      
  }
  </code></pre>
  
  <p>*Pleade let me know if I'm doing something really wrong</p>