递归承诺不返回

Question

我有一个这样的递归函数

function missingItemsPromise() {
    return new Promise(resolve => {
        if (missingItems == 0) {
            console.log('resolves');
            console.log(products);
            return resolve();
        } else {
            page++;
            url = getUrl(id, page);
            http.get(url, function(xres) {
                xres.setEncoding('utf8');
                xres.on('data', function (xtraBody) {
                    console.log('calling');
                    var xtraJson = JSON.parse(xtraBody);
                    var xtraProducts = xtraJson['products'];
                    products = products.concat(xtraProducts);
                    productsLength = products.length;
                    missingItems = total - productsLength;
                    missingItemsPromise();
                });
            });
        } 
    });
};

我正在像使用它一样使用它

getInitial.
then(missingItemsPromise).
then(() => {
 console.log('hello');   
});

我注意到 hello 永远不会返回，因为我怀疑我在递归调用中创建了多个 Promise，但我不确定如何返回。

如何返回每个递归创建的承诺？

编辑：

function missingItemsPromise() {
    return new Promise(resolve => {
        if (missingItems == 0) {
            console.log('resolves');
            return resolve();
        } else {
            page++;
            url = getUrl(id, page);
            http.get(url, function(xres) {
                xres.setEncoding('utf8');
                xres.on('data', function (xtraBody) {
                    console.log('calling');
                    var xtraJson = JSON.parse(xtraBody);
                    var xtraProducts = xtraJson['products'];
                    products = products.concat(xtraProducts);
                    productsLength = products.length;
                    missingItems = total - productsLength;
                    missingItemsPromise();
                    resolve();
                });
            });
        }
    });
};

结果

calling
hello <----notice here that it's already resolving once the first call resolve 
is called
calling
calling
resolves

Answer 1

递归是一种函数式遗产，因此将其与函数式风格一起使用会产生最佳效果。这意味着编写函数来接受和操作其输入（而不是依赖于外部状态）和返回值（而不是依赖于突变或副作用）。

另一方面，您的程序调用不带参数的函数，使用外部状态 missingItems、products、productsLength、total、page 并使用像 page++ 这样的突变和像 @987654333 这样的重新分配@、productsLength = ...、missingItems = ...。我们会解决所有这些问题！

我只是想通过这个来爆破，希望它能让你走上正确的道路。如果您在最后遇到困难，我会链接一些其他答案，这些答案更详细地解释了此处使用的技术。

const getAllProducts = async (page = 0) =>
  asyncUnfold
    ( async (next, done, [ res, nextPage ]) =>
      res.products.length === 0
          ? done ()
          : next ( res.products                               // value to add to output
                 , [ await getPage (nextPage), nextPage + 1 ] // next state
                 )
    , [ await getPage (page), page + 1 ] // initial state
    )

我们介绍上面我们使用的getPage helper

const getPage = async (page = 0, itemsPerPage = 5) =>
  getProducts (page * itemsPerPage, itemsPerPage)
    .then (res => res.json ())

接下来，出于本演示的目的，我们介绍了一个假的getProducts 函数和一个假的DB，其中每个产品只是一个数字。我们还使用delay 来模拟真实的网络延迟。

在您的实际程序中，您只需要提供一个getProducts 函数，可以使用offset 和limit 输入查询产品

// fakes used for demonstration below
const getProducts = (offset = 0, limit = 1) =>
  Promise.resolve
    ({ json: () =>
        ({ products: DB.slice (offset, offset + limit) })
    })
  .then (delay)

const delay = (x, ms = 250) =>
  new Promise (r => setTimeout (r, ms, x))

const DB = 
  [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
  , 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
  , 21, 22, 23, 24, 25, 26, 27, 28, 29, 30
  , 31, 32, 33
  ]

下面我们演示运行程序。 getAllProducts 是一个熟悉的异步函数，它返回一个 Promise 的结果。我们链接了一个.then 调用，因此我们可以在控制台中看到所有产品页面输出

getAllProducts () .then (console.log, console.error)
// ~2 seconds later
// [ [ 1, 2, 3, 4, 5 ]
// , [ 6, 7, 8, 9, 10 ]
// , [ 11, 12, 13, 14, 15 ]
// , [ 16, 17, 18, 19, 20 ]
// , [ 21, 22, 23, 24, 25 ]
// , [ 26, 27, 28, 29, 30 ]
// , [ 31, 32, 33 ]
// ]

与其按页面对产品进行分组，不如将所有产品返回到一个数组中。我们可以稍微修改getAllProducts 来实现这一点

const concat = (xs, ys) =>
  xs .concat (ys)

const concatAll = (arrays) =>
  arrays .reduce (concat, [])

const getAllProducts = async (page = 0) =>
  asyncUnfold
    ( ... )
    .then (concatAll)

getAllProducts () .then (console.log, console.error)
// ~2 seconds later
// [ 1, 2, 3, 4, 5, 6, 7, ..., 31, 32, 33 ]

最后，我们介绍asyncUnfold

const asyncUnfold = async (f, initState) =>
  f ( async (value, nextState) => [ value, ...await asyncUnfold (f, nextState) ]
    , async () => []
    , initState
    )

完整程序演示

// dependencies -------------------------------------------------
const asyncUnfold = async (f, initState) =>
  f ( async (value, nextState) => [ value, ...await asyncUnfold (f, nextState) ]
    , async () => []
    , initState
    )

const concat = (xs, ys) =>
  xs .concat (ys)
  
const concatAll = (arrays) =>
  arrays .reduce (concat, [])
  

// fakes --------------------------------------------------------
const getProducts = (offset = 0, limit = 1) =>
  Promise.resolve
    ({ json: () =>
        ({ products: DB.slice (offset, offset + limit) })
    })
  .then (delay)

const delay = (x, ms = 250) =>
  new Promise (r => setTimeout (r, ms, x))

const DB = 
  [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
  , 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
  , 21, 22, 23, 24, 25, 26, 27, 28, 29, 30
  , 31, 32, 33
  ]

// actual program
const getAllProducts = async (page = 0) =>
  asyncUnfold
    ( async (next, done, [ res, nextPage ]) =>
      res.products.length === 0
          ? done ()
          : next ( res.products
                 , [ await getPage (nextPage), nextPage + 1 ]
                 )
    , [ await getPage (page), page + 1 ]
    )
    .then (concatAll)
    
const getPage = async (page = 0, itemsPerPage = 5) =>
  getProducts (page * itemsPerPage, itemsPerPage)
    .then (res => res.json ())

// demo ---------------------------------------------------------
getAllProducts ()
  .then (console.log, console.error)

// ~2 seconds later
// [ 1, 2, 3, ..., 31, 32, 33 ]

我已经回答的关于递归和承诺的其他问题

• Recursion call async func with promises gets Possible Unhandled Promise Rejection
• Recursive JS function with setTimeout
• Promise with recursion

异步和递归是不同的概念。如果您正在为asyncUnfold 苦苦挣扎，那么首先了解它的同步对应物unfold 可能会有所帮助。这些问答可能有助于区分两者。

• Functional way to create an array of numbers
• Loop until... with Ramda
• Node.js recursively list full path of files

Answer 2

您缺少 else 条件中的 return 语句

function missingItemsPromise() {
    return new Promise(resolve => {
        if (missingItems == 0) {
            console.log('resolves');
            console.log(products);
            return resolve();
        } else {
            page++;
            url = getUrl(id, page);
            http.get(url, function(xres) {
                xres.setEncoding('utf8');
                xres.on('data', function (xtraBody) {
                    console.log('calling');
                    var xtraJson = JSON.parse(xtraBody);
                    var xtraProducts = xtraJson['products'];
                    products = products.concat(xtraProducts);
                    productsLength = products.length;
                    missingItems = total - productsLength;
                   return  missingItemsPromise(); //this is the line that changes
                });
            });
        } 
    });
};