Question

我需要在NodeJS中从后端调用第三方api并将数据返回到前端的ajax调用

以下是我的代码：

router.post('/get_data', function(request, response){
var city_name = request.body.city_name;
if(city_name in city_name_done){

}
else {
    city_name_done.push(city_name);
    console.log('city_name: ' + city_name);
    var options = {
        host : 'api.openweathermap.org',
        path : '/data/2.5/forecast/daily?q=' + city_name + '&mode=json&units=metric&cnt=14&appid=75e843de569fb57a783c2e73fd9a7bb5',
        method : 'GET'
    }
    var maybe = '';
    console.log('till here')
    var req = http.request(options, function(res){
        var body = "";
        res.on('data', function(data) {
            console.log('data came');
            body += data;
        });
        res.on('end', function() {
            console.log('ended too');
            maybe = JSON.parse(body);
            console.log(maybe.city);
        });
    });
    console.log('here too man');
    req.on('error', function(e) {
        console.log('Problem with request: ' + e.message);
    });
    response.send(maybe);
}
});

我可以从前端的ajax post请求中获取city_name参数但是每次执行post请求时我都会收到500内部服务器错误

PS：请原谅我的英语甚至问题的级别因为我是NodeJS的绝对初学者

Answer 1

您正在从函数调用之外返回响应，该响应从第三方API获取数据。您需要从res.on('end'部分返回回复。这是因为api调用是异步的，我们必须等待响应。

res.on('end', function() {
     console.log('ended too');
     maybe = JSON.parse(body);
     console.log(maybe.city);
     response.send(maybe);
});

完整的代码是

    router.post('/get_data', function(request, response){
var city_name = request.body.city_name;
if(city_name in city_name_done){

}
else {
    city_name_done.push(city_name);
    console.log('city_name: ' + city_name);
    var options = {
        host : 'api.openweathermap.org',
        path : '/data/2.5/forecast/daily?q=' + city_name + '&mode=json&units=metric&cnt=14&appid=75e843de569fb57a783c2e73fd9a7bb5',
        method : 'GET'
    }
    var maybe = '';
    console.log('till here')
    var req = http.request(options, function(res){
        var body = "";
        res.on('data', function(data) {
            console.log('data came');
            body += data;
        });
        res.on('end', function() {
            console.log('ended too');
            maybe = JSON.parse(body);
            console.log(maybe.city);
            response.send(maybe);
        });
    });
    console.log('here too man');
    req.on('error', function(e) {
        console.log('Problem with request: ' + e.message);
    });

}
});

Answer 2

使用Postman到api.openweathermap.org/data/2.5/forecast/daily?q=Miami&mode=json&units=metric&cnt=14&appid=75e843de569fb57a783c2e73fd9a7bb5做一个简单的'GET'确实会返回结果。

所以看起来你可能会遗漏像var http = require('http');这样的东西：）

Answer 3

我强烈建议您使用request中的npm模块，使请求更容易处理。请参阅this。

首先：此处不需要maybe。您从API中获取JSON字符串，然后将其解析为对象，但是当您发送它时，您需要再次对其进行字符串化。所以只需坚持字符串形式的传入数据。

其次：在response.send事件之外调用end：所以发生的事情是 - JavaScript是异步的 - 它将调用API，但不等待响应返回，因为您自己已经向您的客户发送了回复。您需要将其放入end回调中，以便实际等待API。

所以这就是我要做的事情：

router.post('/get_data', function(request, response){
var city_name = request.body.city_name;
if(city_name in city_name_done){
    return;
}
else {
    city_name_done.push(city_name);
    console.log('city_name: ' + city_name);
    var options = {
        host : 'api.openweathermap.org',
        path : '/data/2.5/forecast/daily?q=' + city_name + '&mode=json&units=metric&cnt=14&appid=75e843de569fb57a783c2e73fd9a7bb5',
        method : 'GET'
    }
    console.log('till here')
    var req = http.request(options, function(res){
        var body = "";
        res.on('data', function(data) {
            console.log('data came');
            body += data;
        });
        res.on('end', function() {
            console.log('ended too');
            console.log(maybe.city);
            response.send(body);
        });
    });
    console.log('here too man');
    req.on('error', function(e) {
        console.log('Problem with request: ' + e.message);
    });
}
});

NodeJS - 来自app.js的第三方Api电话

3 个答案: