我正在构建一个愚蠢的音乐问答游戏进行学习。我需要用来自deezer api的相关音乐填充视图。
我需要什么:
因此,I found my way直到第3步
但是我无法找到如何正确地发送相同请求4次(对于每个艺术家),到目前为止,我的研究没有给我任何帮助
function deezer() {
const reqGenero = new Request('https://api.deezer.com/genre');
fetch(reqGenero)
.then(response => {
if (response.status === 200) {
return response.json();
} else {
throw new Error('Erro ao pegar gêneros');
}
})
.then(generos => {
/* 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 => {
if (response.status === 200) {
return response.json();
} else {
throw new Error('Erro ao pegar artistas');
}
})
.then(artistas => {
/* 1 música de 4 artistas */
var artistasIds = [];
for(var i = 0; i <= 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 => {
console.error(error);
});
}
*请让我知道我做错什么了
答案 0 :(得分:2)
如果显式使用诺言(请参见下面的async
函数),我可能会这样处理;请参阅***
条注释以获取解释:
// *** Give yourself a helper function so you don't repeat this logic over and over
function fetchJson(errmsg, ...args) {
return fetch(...args)
.then(response => {
if (!response.ok) { // *** .ok is simpler than .status == 200
throw new Error(errmsg);
}
return response.json();
});
}
function deezer() {
// *** Not sure why you're using Request here?
const reqGenero = new Request('https://api.deezer.com/genre');
fetchJson('Erro ao pegar gêneros', reqGenero)
.then(generos => {
/* pega genero aleatorio */
var generoId = generos.data[Math.floor(Math.random() * 10 + 1)].id;
//console.log('\ngenero... ' + generoId);
return fetchJson('Erro ao pegar artistas', 'https://api.deezer.com/genre/' + generoId + '/artists')
})
.then(artistas => {
/* 1 música de 4 artistas */
// *** Use Promise.all to wait for the four responses
return Promise.all(artistas.data.slice(0, 4).map(
entry => fetchJson('Erro ao pegar música', 'https://api.deezer.com/artist/' + entry.id + '/top')
));
})
.then(musica => {
// *** Use musica here, it's an array of the music responses
})
.catch(error => {
console.error(error);
});
}
假设您要使用deezer
中的结果。如果您想deezer
返回结果 (四首歌曲的承诺),则:
// *** Give yourself a helper function so you don't repeat this logic over and over
function fetchJson(errmsg, ...args) {
return fetch(...args)
.then(response => {
if (!response.ok) { // *** .ok is simpler than .status == 200
throw new Error(errmsg);
}
return response.json();
});
}
function deezer() {
const reqGenero = new Request('https://api.deezer.com/genre');
return fetchJson('Erro ao pegar gêneros', reqGenero) // *** Note the return
.then(generos => {
/* pega genero aleatorio */
var generoId = generos.data[Math.floor(Math.random() * 10 + 1)].id;
//console.log('\ngenero... ' + generoId);
return fetchJson('Erro ao pegar artistas', 'https://api.deezer.com/genre/' + generoId + '/artists')
})
.then(artistas => {
/* 1 música de 4 artistas */
// *** Use Promise.all to wait for the four responses
return Promise.all(artistas.data.slice(0, 4).map(
entry => fetchJson('Erro ao pegar música', 'https://api.deezer.com/artist/' + entry.id + '/top')
));
});
// *** No `then` using the results here, no `catch`; let the caller handle it
}
第二个版本的async
函数版本:
// *** Give yourself a helper function so you don't repeat this logic over and over
async function fetchJson(errmsg, ...args) {
const response = await fetch(...args)
if (!response.ok) { // *** .ok is simpler than .status == 200
throw new Error(errmsg);
}
return response.json();
}
async function deezer() {
const reqGenero = new Request('https://api.deezer.com/genre');
const generos = await fetchJson('Erro ao pegar gêneros', reqGenero);
var generoId = generos.data[Math.floor(Math.random() * 10 + 1)].id;
//console.log('\ngenero... ' + generoId);
const artistas = await fetchJson('Erro ao pegar artistas', 'https://api.deezer.com/genre/' + generoId + '/artists');
/* 1 música de 4 artistas */
// *** Use Promise.all to wait for the four responses
return Promise.all(artistas.data.slice(0, 4).map(
entry => fetchJson('Erro ao pegar música', 'https://api.deezer.com/artist/' + entry.id + '/top')
));
}
答案 1 :(得分:1)
您可以使用Promise#all创建4个请求,并等待所有请求完成 。
.then(artistas => {
/* 1 música de 4 artistas */
const artistasPromises = artistas.data.map(artista =>
fetch("https://api.deezer.com/artist/" + artista.id + "/top").catch(
err => ({ error: err })
)
);
return Promise.all(artistasPromises);
}).then(musicList => {
console.log(musicList);
});
请注意catch()
。这样可以确保即使提取失败,其他提取结果也不会被忽略。这是因为Promise#all的工作方式。因此,您需要遍历musicList
并检查是否存在任何形状为{ error: /* error object */ }
的对象,并在处理列表时将其忽略。
答案 2 :(得分:1)
您可以替换语句
// CAN I SEND THIS REQUEST 4 TIMES?
return fetch('https://api.deezer.com/artist/' + ids + '/top');
使用
const fetchResults = [];
artistasIds.forEach(function(ids){
fetchResults.push(fetch('https://api.deezer.com/artist/' + ids + '/top'));
});
return Promise.all(fetchResults);
在然后条件下,您将获得包含每个艺术家的顶级音乐的值数组。我没有使用给定的API进行检查,但理想情况下应该可以使用。
答案 3 :(得分:1)
是的,您可以发出5个请求(而不是4个0-4),然后等待每个请求完成。 使用Array.prototype.map创建一个请求承诺数组。(优先于for-forEach和array.push)
和Promise.all 等待所有的诺言完成,如果没有失败,它将返回已解决的响应数组。
.then(artistas => {
/* 1 música de 4 artistas */
var artistasIds = [];
let ids = artistas.data.map(artist => artist.id).slice(0, 4);
requests = ids.map(id => fetch(`https://api.deezer.com/artist/${id}/top`));
return Promise.all(requests);
}
})