Question

我有一个脚本应该尝试连接3次到服务器，如果失败导航用户到No Connection page。

我不知道如何尝试连接3次，然后如果失败则将用户导航到No Connection page。

如何使用我的代码来解决此问题。

这是我到目前为止的代码：

const fetchModule = require('fetch');
const frameModule = require('ui/frame');

let timeTrig;
const waitPeriod = 7000;

function checkServerStatus() {
    if (timeTrig) {
        clearTimeout(timeTrig);
        timeTrig = null;
    }
    timeTrig = setTimeout(() => {
    frameModule.topmost().navigate('./views/no-connection');
    }, waitPeriod);
    return fetchModule.fetch(`${randomUrl}auth/ping`)
        .then((result) => {
            clearTimeout(timeTrig);
            if (result && result.ok) {
                return true;
            }
            frameModule.topmost().navigate({
                moduleName: views/no-connection,
                transition: {
                    name: 'slide',
                },
            });
        })
        .catch(() => {
            if (timeTrig) {
                clearTimeout(timeTrig);
                timeTrig = null;
            }
            frameModule.topmost().navigate({
                moduleName: 'views/no-connection',
                transition: {
                    name: 'slide',
                },
            });
        });
}

感谢任何帮助。

Answer 1

也许这样的东西就是你想要的 - 不确定你的代码的哪一部分应该被重试，但这可能会给你一些想法

const fetchModule = require('fetch');
const frameModule = require('ui/frame');

const waitPeriod = 7000;

// retry "cont" times or until the promise resolves
const retry = (cont, fn) => fn().catch(err => cont > 0 ? retry(cont - 1, fn) : Promise.reject(err));

// failure code - reason will be the last failure reason ('not ok' or 'timeout') - not that it is used in your code
const noConnection = reason => frameModule.topmost().navigate({
    moduleName: 'views/no-connection',
    transition: {
        name: 'slide',
    },
});

// lets race between (resolved or rejected) fetch and a rejected timeout
const doFetch = () => Promise.race([
    fetchModule.fetch(`${randomUrl}auth/ping`).then(result => {
        if (result && result.ok) {
            return true;
        }
        return Promise.reject('not ok');
    }),
    // above can be written as
    // fetchModule.fetch(`${randomUrl}auth/ping`).then(result => (result && result.ok) || Promise.reject('not ok')),
    new Promise((resolve, reject) => setTimeout(reject, waitPeriod, 'timeout'))
]);

const checkServerStatus = () => {
    retry(3, doFetch).catch(noConnection)
};

或者，这是相同的代码，只是组织不同。我认为更好：p

const fetchModule = require('fetch');
const frameModule = require('ui/frame');

const waitPeriod = 7000;

// retry "cont" times or until the promise resolves
const retry = (cont, fn) => fn().catch(err => cont > 0 ? retry(cont - 1, fn) : Promise.reject(err));
// reject after a given timeout
const delayedReject = (delay, reason) => new Promise((_, reject) => setTimeout(reject, delay, reason));
//
const doFetch = () => fetchModule.fetch(`${randomUrl}auth/ping`)
.then(result => (result && result.ok) || Promise.reject('not ok'));

const checkServerStatus = () => retry(3, () => Promise.race([doFetch(), delayedReject(waitPeriod, 'timeout')]))
.catch(reason => frameModule.topmost().navigate({
    moduleName: 'views/no-connection',
    transition: {
        name: 'slide',
    },
}));

Answer 2

这＆＃34;尝试n次然后失败＆＃34;问题的类型经常出现，以http提取或网络套接字的形式出现，什么不是。我之前实际上已经实现了promise TryNTimes功能。

这是一个更有趣的版本，我认为非常优雅和可重复使用。

我们有几个实用程序函数来模仿获取请求和承诺排序。

以下bring函数就像获取一样，有50％的解析概率。

bring    = (url,i) => new Promise((v,x) => Math.random() > 0.50 ? v(`Data from ${url}`) : x(`Error @ promise # ${i}`))

接下来，我们设置一个大小为n bring个函数的数组。

promises = (f,n) => Array(n).fill(f),

接下来，我们对递归bring函数的promise进行递归排序。如果被调用的bring函数返回的promise会解析，我们会返回该分辨率，但如果拒绝，那么我们继续执行数组中的下一个，直到不再剩下。显然，数组的大小是尝试计数。

＆＃13;

var bring    = (url,i) => new Promise((v,x) => Math.random() > 0.50 ? v(`Data from ${url}`) : x(`Error @ promise # ${i}`)),
    promises = (f,n) => Array(n).fill(f),
    tryProms = ([f,...fs], url, i = 1) => f(url,i).then(v => `${v} @ try # ${i}`,
                                                        e => (console.log(`Failed @ try #${i}`),
                                                              fs.length ? tryProms(fs,url,++i)
                                                                        : `All ${i} tries are failed`));
tryProms(promises(bring,3),"http://www.someAPI/data").then(console.log, console.log);

＆＃13;

循环承诺，链承诺

2 个答案: