我制作了这个虚拟代码,以更好地理解promise的工作方式,模仿了我必须“承诺化”的更复杂的软件。 在所附的代码中,我希望事件触发并按以下顺序记录:
但是,如您所见,如果在运行它,则在步骤2和3之间将打印“ after”字符串。 显然,我在处理异步逻辑时一定做错了。 感谢您的帮助!
const obj = {
"rows": [{
"type": "A",
"value": 0
}, {
"type": "B",
"value": 0
}, {
"type": "C",
"value": 0
}]
}
let promises = [];
function delay(ms) {
return new Promise(resolve => setTimeout(resolve, ms));
}
const alter_value = async(row, to_add, time) => {
await delay(time);
row.value = row.value + to_add;
console.log("done with " + row.type);
return true;
}
const two = async() => {
obj.rows.forEach(async(row) => {
switch (row.type) {
case "A":
console.log("A detected");
promises.push(alter_value(row, 1, 1000))
promises.push(alter_value(row, 2, 1800))
break;
case "B":
console.log("B detected");
promises.push(alter_value(row, 5, 1400))
break;
case "C":
console.log("C detected");
promises.push(alter_value(row, 200, 2400))
break;
}
});
return promises;
}
const one = async() => {
console.log("before");
Promise.all(two()).then(console.log("after"));
}
one();
答案 0 :(得分:1)
我看到您的代码至少有两个问题,解释了您得到的结果:
two
函数不应为async
。 async
函数返回隐式Promise
。在这里,您只想返回一个已经构造好的Promise
数组,因此您需要一个普通函数。.then(console.log("after"))
将立即执行console.log
:then()
期望某个函数稍后执行,因此您必须将其更改为.then(() => console.log("after"))
。它变成:
const obj = {
"rows": [{
"type": "A",
"value": 0
}, {
"type": "B",
"value": 0
}, {
"type": "C",
"value": 0
}]
};
function delay(ms) {
return new Promise(resolve => setTimeout(resolve, ms));
}
const alter_value = async (row, to_add, time) => {
await delay(time);
row.value = row.value + to_add;
console.log("done with " + row.type);
return true;
};
const two = () => {
const promises = [];
obj.rows.forEach(async(row) => {
switch (row.type) {
case "A":
console.log("A detected");
promises.push(alter_value(row, 1, 1000));
promises.push(alter_value(row, 2, 1800));
break;
case "B":
console.log("B detected");
promises.push(alter_value(row, 5, 1400));
break;
case "C":
console.log("C detected");
promises.push(alter_value(row, 200, 2400));
break;
}
});
return promises;
};
const one = async () => {
console.log('before');
Promise.all(two()).then(() => console.log('after'));
};
one();
请注意,作为.then()
的替代方案,您也可以在await
上简单地使用Promise.all
,以使代码更加一致:
await Promise.all(two());
console.log('after');
答案 1 :(得分:0)
OP似乎要求同步行为。
演示大纲
函数modVal(i, value, time)
参数是从异步函数syncro()
参数派生的:const sync
对象数组。每个对象都包含obj.rows
的索引,该索引obj.rows[i].value
的值以及modVal()
的Promise中的超时时间。
参数和参数: const sync = [{r: 0, v: 1, t: 1000}, ...];
seq.r::obj.rows[
数字 ]
seq.v: obj.rows[seq.r].value +=
编号
seq.t: ...resolve(obj.rows[i].value += value),
数字);
sync
数组通过for...of
循环进行迭代。在每次迭代中,await modVal()
被同步调用。
let obj = {
"rows": [{
"type": "A",
"value": 0
}, {
"type": "B",
"value": 0
}, {
"type": "C",
"value": 0
}]
}
const sync = [{
r: 0,
v: 1,
t: 1000
}, {
r: 0,
v: 2,
t: 1800
}, {
r: 1,
v: 5,
t: 1400
}, {
r: 2,
v: 200,
t: 2400
}];
const syncro = async(sync) => {
const modVal = (i, value, time) => {
return new Promise(resolve => {
setTimeout(() => resolve(obj.rows[i].value += value), time);
});
}
for (let seq of sync) {
await modVal(seq.r, seq.v, seq.t);
console.log(JSON.stringify(obj.rows));
}
}
syncro(sync);
.as-console-row.as-console-row::after {
content:'';
padding:0;
margin:0;
border:0;
width:0;
}