温馨提示:本文翻译自stackoverflow.com,查看原文请点击:android - How to grab json file from external url(PLACES API) using firebase functions

android - 如何使用Firebase函数从外部URL(PLACES API)抓取JSON文件

发布于 2020-03-27 11:00:00

因此,我想使用云功能将请求从我的应用程序发送到Firebase,然后处理进程url并从place api发送回JSON文件

我已经完成/已经完成的工作>在控制台中设置项目并获取Firebase CLI之后,按如下方式创建了云功能

在您发表评论后,这是我的全部功能代码:

const functions = require('firebase-functions');
const admin = require('firebase-admin');
admin.initializeApp();
const rp = require('request-promise');

exports.fetch = functions.https.onCall((req, res) => {

    const url = req.url + '&key=MY_API_KEY';
    var options = {
        uri: url, // Automatically parses the JSON string in the response
        json: true
    };  

    rp(options)
    .then(result => {
        console.log('Get response:' + response.statusCode);
        return res.type('application/json').send(result);
    }).catch(err => {
        // API call failed...
        return res.send({'Error': err});
    });

})

在java类中传递这样的值

     private Task<String> addMessage(String url) {
            // Create the arguments to the callable function.
            Map<String, Object> data = new HashMap<>();
            data.put("url", url);///PASSING VALUES HERE
            return mFunctions
                    .getHttpsCallable("fetch")
                    .call(data)
                    .continueWith(task -> 
(String) Objects.requireNonNull(task.getResult()).getData());
        }

现在,我的问题是在使用Firebase CLI部署新功能代码时16:8 error Each then() should return a value or throw promise/always-return,出现错误

谁能指导我..,

URL将是这样的: https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=17.4369681,78.4473887&radius=5000&type=airport&sensor=true&key=MY_KEY

这是来自控制台的日志详细信息

2019-07-05T10:06:35.025308453Z D fetch: Function execution started
2019-07-05T10:06:35.265840608Z D fetch: Function execution took 241 ms, finished with status code: 200
2019-07-05T10:06:45.162Z I fetch: Get response:undefined
2019-07-05T10:06:46.062Z E fetch: Unhandled rejection
2019-07-05T10:06:46.062Z E fetch: TypeError: res.send is not a function
    at rp.then.catch.err (/srv/index.js:22:14)
    at bound (domain.js:301:14)
    at runBound (domain.js:314:12)
    at tryCatcher (/srv/node_modules/bluebird/js/release/util.js:16:23)
    at Promise._settlePromiseFromHandler (/srv/node_modules/bluebird/js/release/promise.js:517:31)
    at Promise._settlePromise (/srv/node_modules/bluebird/js/release/promise.js:574:18)
    at Promise._settlePromise0 (/srv/node_modules/bluebird/js/release/promise.js:619:10)
    at Promise._settlePromises (/srv/node_modules/bluebird/js/release/promise.js:695:18)
    at _drainQueueStep (/srv/node_modules/bluebird/js/release/async.js:138:12)
    at _drainQueue (/srv/node_modules/bluebird/js/release/async.js:131:9)
    at Async._drainQueues (/srv/node_modules/bluebird/js/release/async.js:147:5)
    at Immediate.Async.drainQueues (/srv/node_modules/bluebird/js/release/async.js:17:14)
    at runCallback (timers.js:810:20)
    at tryOnImmediate (timers.js:768:5)
    at processImmediate [as _immediateCallback] (timers.js:745:5)

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提问者
trinadh thatakula
被浏览
122
trinadh thatakula 2019-07-05 20:10

这对我有用。在promise之前使用return返回promise的结果,并且在使用returnfunctions.https.onCall使用,我们需要res.send使用functions.https.onRequest它花了我三天的时间,并且不需要json=true在URL返回json时添加请求承诺选项,这使事情变得复杂

const functions = require('firebase-functions');
const admin = require('firebase-admin');
admin.initializeApp();
const rp = require('request-promise');

exports.fetch = functions.https.onCall((req, res) => {

    const url = req.url + '&key=MY_API_KEY';
    var options = {
        uri: url, // Automatically parses the JSON string in the response
    };

    return rp(options)
    .then(result => {
        console.log('here is response: ' + result);
        return result;
    }).catch(err => {
        // API call failed...
        return err;
    });

})

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