如何嵌套承诺（Promise.all 内的 Promise.all）

Question

考虑以下嵌套Promises结构：

 const getData = async() => { const refs = [{ name: "John33", age: 33 }, { name: "John34", age: 34 }, { name: "John35", age: 35 }, { name: "John36", age: 36 } ]; let source = [{ name: "John30", age: "unknown" }, { name: "John31", age: "unknown" }, { name: "John32", age: "unknown" }, { name: "John33", age: "unknown" }, { name: "John34", age: "unknown" }, { name: "John35", age: "unknown" }, { name: "John36", age: "unknown" }, { name: "John37", age: "unknown" }, { name: "John38", age: "unknown" }, { name: "John39", age: "unknown" } ]; const resolver = doc => { return new Promise(doc => { let clone = { ...doc }; let found = refs.find(ref => { return ref.name === doc.name; }); if (found) clone.age = found.age; return clone; }); }; let getRefs = (doc, refs) => { const promises = refs.map(r => { resolver(doc).then(result => { return result; }); }); return Promise.all(promises); }; let getCursorData = (cursor, refs, data) => { const promises = cursor.forEach(doc => { console.log("Getting cursor for " + doc.name); let clone = { ...doc }; return getRefs(clone, refs).then(result => { console.log("Getting refs for " + clone.name); data.push(result); }); return; }); return Promise.all(promises); }; // Get data let data = []; await getCursorData(source, refs, data); console.log("Returned data: "); console.log(data); return data; }; console.log("Begin"); getData().then(result => { console.log("End"); console.log(result) });

出于某种原因，我没有到代码的末尾（ End没有被打印）。 我怀疑有一些位置或缺少回报，但我在没有找到解决方案的情况下被卡住了。

我怎样才能使这个代码结构按预期工作，如下所示：

1. 遍历源（我的数据来自数据库
2. 对于每个寄存器，应用引用更改（在示例中更改age ）
3. 返回固定引用的数据

此代码的预期结果是使用当前的 promise 结构获取具有固定可用引用的原始数据（ source ）：

[
      name: "John30",
      age: "unknown"
    },
    {
      name: "John31",
      age: "unknown"
    },
    {
      name: "John32",
      age: "unknown"
    },
    {
      name: "John33",
      age: 33
    },
    {
      name: "John34",
      age: 34
    },
    {
      name: "John35",
      age: 35
    }, {
      name: "John36",
      age: 36
    },
    {
      name: "John37",
      age: "unknown"
    },
    {
      name: "John38",
      age: "unknown"
    },
    {
      name: "John39",
      age: "unknown"
    }
]

Answer 1

干得好！

 const SimulatedDatabaseCall = new Promise(resolve => { setTimeout(() => { resolve([ { name: 'John33', age: 33 }, { name: 'John34', age: 34 }, { name: 'John35', age: 35 }, { name: 'John36', age: 36 } ]); }, 200); }); let source = [ { name: 'John30', age: 'unknown' }, { name: 'John31', age: 'unknown' }, { name: 'John32', age: 'unknown' }, { name: 'John33', age: 'unknown' }, { name: 'John34', age: 'unknown' }, { name: 'John35', age: 'unknown' }, { name: 'John36', age: 'unknown' }, { name: 'John37', age: 'unknown' }, { name: 'John38', age: 'unknown' }, { name: 'John39', age: 'unknown' } ]; async function updateSource(source, SimulatedDatabaseCall) { await SimulatedDatabaseCall; SimulatedDatabaseCall.then(_database => { var sourceMap = source.map(_sourceOBJ => { return _sourceOBJ.name; }); _database.forEach(_databaseOBJ => { var index = sourceMap.indexOf(_databaseOBJ.name); source[index].age = _databaseOBJ.age; }); }); return source; } updateSource(source, SimulatedDatabaseCall).then(_val => { console.log(_val); });