我想知道如何从承诺中回报承诺。 E.g。
我有这样的结构:
doAsyncStuff() // a promise
.then( function(res) {
doAnotherAsyncStuff(res) // another promise
.then( makeSomeThings )
.then( function(anotherRes, opts) {
...
})
})
.then( ... )
我想这样写:
doAsyncStuff() // a promise
.then( function(res) {
doAnotherAsyncStuff(res) // another promise
.then( makeSomeThings )
// and somehow push-out promise
})
.then( function(anotherRes) {
...
})
.then( ... )
我怎样才能达到这样的效果?
问题
var Promise = require('bluebird');
//noinspection JSUnresolvedFunction
var bcrypt = Promise.promisifyAll(require('bcrypt'));
var Sequelize = require('sequelize');
var config = require('config');
var sequelize = new Sequelize(config.get('db.connstring'));
//noinspection JSUnresolvedFunction
var User = sequelize.define('user', {
name: {
type: Sequelize.STRING
},
email: {
type: Sequelize.STRING,
validate: {
isEmail: true
}
},
passwordHash: {
type: Sequelize.STRING
},
isConfirmed: {
type: Sequelize.BOOLEAN,
allowNull: false,
defaultValue: false
}
}, {
freezeTableName: true,
classMethods: {
login: Promise.method(function (email, password) {
if (!email || !password) throw new Error('Email and password are both required');
var rv = this
.find({where: {email: email.toLowerCase().trim()}})
.then(function (user) {
return bcrypt.compareAsync(password, user.passwordHash).then(function (res) {
console.log(email, password, res);
});
// if i dont use pacthed compare here, i have no problem ..
// return bcrypt.compare(password, user.passwordHash, function(err, res) {
// console.log(email, password, res);
// });
});
console.log('B', rv);
return rv;
})
}
});
sequelize.sync({force: true}).then(function () {
var pwd = 'pwd';
//noinspection JSUnresolvedFunction
bcrypt.hashAsync(pwd, 4).then(function (salt) {
var u1 = User.create({
name: 'u1',
email: 'u1@ex.com',
passwordHash: salt
}).then(function (result) {
User.login('u1@ex.com', pwd).then(function (res) {
console.log('A', res)
})
});
});
});
答案 0 :(得分:5)
回复你的另一个承诺
doAsyncStuff() // a promise
.then( function(res) {
return doAnotherAsyncStuff(res) // another promise
})
.then( function(anotherRes) {
...
})
.then( ... )
答案 1 :(得分:2)
如果一个函数表示将前一个(已解决的)结果转换为下一个结果或一个将解析为下一个结果的promise,则只需将其传递给{{1直接。
then
答案 2 :(得分:0)
正如其他答案已经建议的那样,您必须从return
回调中 then
。无论你返回一个简单的价值还是一个价值的承诺都没关系,两者都有效;但是当你什么也没有返回时,结果的承诺将通过undefined
来解决。
在您的特定示例中,它位于:
doAsyncStuff() // a promise
.then( function(res) {
return doAnotherAsyncStuff(res) // another promise
// ^^^^^^
.then( makeSomeThings )
…
…
.then(function (user) {
return bcrypt.compareAsync(password, user.passwordHash).then(function (res) {
console.log(email, password, res);
return res;
// ^^^^^^
});
在您的上一个示例中,当您使用return
时,您甚至可以unnest拨打电话:
var pwd = 'pwd';
sequelize.sync({force: true}).then(function () {
//noinspection JSUnresolvedFunction
return bcrypt.hashAsync(pwd, 4);
}).then(function (salt) {
return User.create({
name: 'u1',
email: 'u1@ex.com',
passwordHash: salt
});
}).then(function (result) {
return User.login('u1@ex.com', pwd);
}).then(function (res) {
console.log('A', res);
});