使用async/await和Promise.all循环调用递归函数

Question

我有一个用例，我试图循环遍历对象数组，其中我需要做出一些GraphQL请求，这些请求可能对数组中的给定对象有一些分页。我试图通过将递归函数推到承诺数组来提高性能，然后使用Promse.all来解决所有这些问题。

我遇到了一个问题，虽然我从Promise.all得到undefined响应-最终目标是对数组中的每个唯一对象有以下响应:

[{
account: test1,
id: 1,
high: 2039,
critical: 4059
},
{
account: test2,
id: 2,
high: 395,
critical: 203
}]

…在递归完成分页/为给定帐户对象发出所有请求后，我只返回anAccount对象。

下面是示例代码:

const fetch = require('isomorphic-fetch');
const API_KEY = '<key>';
async function main() {
let promises = [];
let accounts = [{'name': 'test1', 'id': 1}, {'name': 'test2' , 'id': 2}];
for (const a of accounts) {
let cursor = null;
let anAccountsResults = [];
promises.push(getCounts(a, anAccountsResults, cursor));
}
let allResults = await Promise.all(promises);
console.log(allResults);
}
async function getCounts(acct, results, c) {
var q = ``;
if (c == null) {
q = `{
actor {
account(id: ${acct.id}) {
aiIssues {
issues(filter: {states: ACTIVATED}) {
issues {
issueId
priority
}
nextCursor
}
}
}
}
}`
} else {
q = `{
actor {
account(id: ${acct.id}) {
aiIssues {
issues(filter: {states: ACTIVATED}, cursor: "${c}") {
issues {
issueId
priority
}
nextCursor
}
}
}
}
}`
}
const resp = await fetch('https://my.api.com/graphql', {
method: 'POST',
headers: {
'Content-Type': 'application/json',
'API-Key': API_KEY
},
body: JSON.stringify({
query: q,
variables: ''}),
});
let json_resp = await resp.json();
let aSingleResult = json_resp.data.actor.account.aiIssues.issues.issues;
let nextCursor = json_resp.data.actor.account.aiIssues.issues.nextCursor;
console.log(nextCursor);
if (nextCursor == null) {
results = results.concat(aSingleResult);
} else {
results = results.concat(aSingleResult);
await getCounts(acct, results, nextCursor);
}
let criticalCount = results.filter(i => i.priority == 'CRITICAL').length;
let highCount = results.filter(i => i.priority == 'HIGH').length;
let anAccount = {
account: acct.name,
id: acct.id,
high: highCount,
critical: criticalCount
};
return anAccount;
}
main();

函数getCounts

中的logginganAccount具有正确的细节，但是当返回它时，记录Promise.all(promises)的输出将生成undefined。是否有更好的方法来处理这种方式，我仍然可以异步运行多个递归函数在循环内并行与Promise.all?

Answer 1

您的主要问题似乎是results = results.concat(aSingleResult);不会改变您传递的数组，而只是在函数内部重新分配局部变量结果，因此anAccount只会使用当前调用中的aSingleResult。与其将内容收集到传递给a参数的结果数组中，不如让每次调用都返回一个新数组。然后在递归的await getCounts(acct, results, nextCursor)调用中，不要忽略返回值。

async function main() {
let promises = [];
const accounts = [{'name': 'test1', 'id': 1}, {'name': 'test2' , 'id': 2}];
const promises = accounts.map(async acct => {
const results = await getIssues(acct);
const criticalCount = results.filter(i => i.priority == 'CRITICAL').length;
const highCount = results.filter(i => i.priority == 'HIGH').length;
return {
account: acct.name,
id: acct.id,
high: highCount,
critical: criticalCount
};
});
const allResults = await Promise.all(promises);
console.log(allResults);
}

const query = `query ($accountId: ID!, $cursor: IssuesCursor) {
actor {
account(id: $accountId) {
aiIssues {
issues(filter: {states: ACTIVATED}, cursor: $cursor) {
issues {
issueId
priority
}
nextCursor
}
}
}
}
}`;
async function getIssues(acct, cursor) {
const resp = await fetch('https://my.api.com/graphql', {
method: 'POST',
headers: {
'Content-Type': 'application/json',
'API-Key': API_KEY
},
body: JSON.stringify({
query: q,
variables: {
accountId: acct.id,
cursor,
}
}),
});
if (!resp.ok) throw new Error(resp.statusText);
const { data, error } = await resp.json();
if (error) throw new Error('GraphQL error', {cause: error});
const { nextCursor, issues } = data.actor.account.aiIssues.issues;
if (nextCursor == null) {
return issues;
} else {
return issues.concat(await getIssues(acct, nextCursor));
}
}