使用 request() 和回调时，如何在 Azure Function 中不调用 context.done() 两次？

Question

我正在编写一个 Azure 函数，它使用 npm 模块请求调用 Google API。处理 body 的回调需要调用 context.done() 否则 context.res 超出范围（如果我在回调中设置它然后等待在我的脚本末尾通过 context.done()).

问题是我不明白如何处理错误和return对客户端的响应和使用回调。

欢迎对我的代码提出任何意见。我要改进。

var request = require('request');

module.exports = function (context, req) {

    if (req.body) {
        incoming = req.body

        if (incoming.components && incoming.components.city
            && incoming.components.state
            && incoming.components.country) {

            // locality = city
            // administrative_area = state
            var country = incoming.components.country;
            var administrative_area = incoming.components.state;
            var locality = incoming.components.city;

            request.get({
                url: "https://maps.googleapis.com/maps/api/geocode/json?components=country:US|locality:city|administrative_area:CA&key=..."
            }, function (error, response, body) {
                if (error) {
                    context.log(error);
                }

                if (!error && response.statusCode == 200) {
                    context.log(body)
                }
                context.res = {
                    status: 200,
                    body: "asdf"
                };
                context.done();
            });

            context.log("here");
        } else {
            context.res = {
                status: 400,
                body: 'There was data but not the right kind. Try {"components": {"country": "US","city": "city","state": "CA"}'
            };
        }

    }
    else {
        context.res = {
            status: 400,
            body: 'Nothing was received. Try {"components": {"country": "US","city": "city","state": "CA"}'
        };
    }
    context.done();
};

Answer 1

您不应在函数结束时调用 context.done()。仅在获得结果时调用它（即当您检测到错误或在回调内部时）。对您的函数进行一点重构和简化：

module.exports = function (context, req) {
    if (!req.body) {
        context.res = {
            status: 400,
            body: 'Nothing was received. Try {"country": "US"}'
        };
        context.done();
        return;
    }

    if (!req.body.country) {
        context.res = {
            status: 400,
            body: 'There was data but not the right kind. Try {"country": "US"}'
        };
        context.done();
        return;
    }        

    request.get({
        url: "https://www.google.com?q=" + req.body.country
    }, function (error, response, body) {
        if (error) {
            context.log(error);
            context.res = {
                status: 500,
                body: "Unexpected error"
            };
        } else {
            if (response.statusCode == 200) {
                context.log(body)
            }
            context.res = {
                status: 200,
                body: "asdf"
            };
        }
        context.done();
    });
};

使用 request() 和回调时，如何在 Azure Function 中不调用 context.done() 两次？

How to not call context.done() twice in Azure Function when using request() and callback?

javascript

azure

azure-functions