以下是我要运行的代码:
async function verifyExistingUsers(db, users) {
return new Promise((resolve, reject) => {
var companies = []
for (const [index, user] of users.entries()) {
let company = await getUserCompany(db, user)
companies.push(company)
if (index === users.length-1) {
resolve(companies)
}
}
})
}
async function getUserCompany(db, user) {
return new Promise((resolve, reject) => {
db.Company.findAll({
where: {
id: user.id,
}
}).then(company => {
resolve(company)
})
}).catch(error => {
reject()
})
}
我一直收到以下错误:
let companies = await getUserCompany(db, user)
^^^^^
SyntaxError: await is only valid in async function
我不能使用forEach
,因为我需要await
。
JavaScript的新手。我在做什么错了?
答案 0 :(得分:2)
作为我的评论的后续:verifyExistingUsers
没有理由“返还新的诺言”。 Node会将自己的return语句包装在一个promise中,因为该函数是'async'。
这里的原始错误是因为您不能有效地在匿名非异步函数await
中使用((resolve, reject) => {})
。
您不会返回新的Promise,而只是在将数据推入数组后返回所需的变量。通过这样做并将函数声明为“异步”,Node会将您的返回值包装在一个您等待其他地方的承诺中。
async function verifyExistingUsers(db, users) {
var companies = []
for (const [index, user] of users.entries()) {
let company = await getUserCompany(db, user)
companies.push(company)
if (index === users.length-1) {
return companies; //effectively returns a promise that resolves to companies
}
}
}
答案 1 :(得分:1)
如果我正确理解了您的问题,这是您需要的解决方案,那么我也解决了同样的问题,您也可以循环使用它,希望对您有所帮助,如果这不是您的答案,请告诉我更新我的信息。回答
const AsyncFuncion = async () => {
let interval = 2000;
const delayPromise = (data, delayDuration) => {
return new Promise((resolve) => {
setTimeout(() => {
// here you can do your operations on db
resolve();
}, delayDuration)
});
};
try {
const userData = // the data you want from db or you can use http request to get that data
const promises = userData.map((data, index) => delayPromise(data, index * interval));
await Promise.all(promises);
setTimeout(function(){
console.log('done') // after 10 minitues it'll repeate it self you can disable it also
AsyncFuncion()
}, 1000 * 60 * 10)
} catch (e) {
console.error('AsyncFuncion', e);
}
}
AsyncFuncion();