Question

我在nodejs上有一个端点，虽然它是一个承诺，但它不会产生输出，它应该返回输出。这是我的要求：

TegucigalpaData: () => {
        let url = `https://maps.googleapis.com/maps/api/geocode/json?address=${encodeURIComponent('Tegucigalpa, Honduras')}`;
        axios.get(url).then( (response) => {
            if(response.data.status === 'ZERO_RESULTS') {
                throw new Error(`Unable to find the address: ${address}`);
            }
            let lat = response.data.results[0].geometry.location.lat;
            let lng = response.data.results[0].geometry.location.lng;
            let wurl = `https://api.darksky.net/forecast/dc6c0d767d55ffe8f172d4c187852db8/${lat},${lng}?lang=es&units=si`;
            return axios.get(wurl);
        }).then( (response) => {
            let temperature = response.data.currently.temperature;
            let apparentTemperature = response.data.currently.apparentTemperature;
            return { temp: temperature, appTemp: apparentTemperature }
        }).catch( (error) => {
            return "Unable to connect to the API servers";
        });
    }

然后，在我的终端上：

app.get('/weatherData', (req, res) => {
    let data = TegucigalpaData();
    res.send(data);
});

但是，我没有像json那样得到任何数据输出。知道我可能做错了什么吗？

Answer 1

尝试直接在处理程序中发送结果：

app.get('/weatherData', (req, res) => {
    let url = `https://maps.googleapis.com/maps/api/geocode/json?address=${encodeURIComponent('Tegucigalpa, Honduras')}`;
    axios.get(url).then( (response) => {
        if(response.data.status === 'ZERO_RESULTS') {
            throw new Error(`Unable to find the address: ${address}`);
        }
        let lat = response.data.results[0].geometry.location.lat;
        let lng = response.data.results[0].geometry.location.lng;
        let wurl = `https://api.darksky.net/forecast/dc6c0d767d55ffe8f172d4c187852db8/${lat},${lng}?lang=es&units=si`;
        return axios.get(wurl);
    }).then( (response) => {
        let temperature = response.data.currently.temperature;
        let apparentTemperature = response.data.currently.apparentTemperature;
        return res.send({ temp: temperature, appTemp: 
        apparentTemperature })
    }).catch( (error) => {
        return res.send(error)
    });
} 
});

在nodejs上获取端点不会产生任何输出

1 个答案: