简体   繁体   中英

node.js + request => node.js + bluebird + request

I'm trying to understand how to write code with promises. Check my code plz. This is right?

Node.js + request:

request(url, function (error, response, body) {
    if (!error && response.statusCode == 200) {
        var jsonpData = body;
        var json;
        try {
            json = JSON.parse(jsonpData);
        } catch (e) {
            var startPos = jsonpData.indexOf('({');
            var endPos = jsonpData.indexOf('})');
            var jsonString = jsonpData.substring(startPos+1, endPos+1);
            json = JSON.parse(jsonString);
        }
        callback(null, json);
    } else {
        callback(error);
    }
});

Node.js + bluebird + request:

request.getAsync(url)
   .spread(function(response, body) {return body;})
   .then(JSON.parse)
   .then(function(json){console.log(json)})
   .catch(function(e){console.error(e)});

How to check response status? I should use if from first example or something more interesting?

You can simply check if the response.statusCode is not 200 in the spread handler and throw an Error from that, so that the catch handler will take care of it. You can implement it like this

var request = require('bluebird').promisifyAll(require('request'), {multiArgs: true});

request.getAsync(url).spread(function (response, body) {
    if (response.statusCode != 200)
        throw new Error('Unsuccessful attempt. Code: ' + response.statusCode);
    return JSON.parse(body);
}).then(console.log).catch(console.error);

And if you notice, we return the parsed JSON from the spread handler, because JSON.parse is not an async function, so we don't have to do it in a separate then handler.

One way to check the status code:

.spread(function(response, body) {
  if (response.statusCode !== 200) {
    throw new Error('Unexpected status code');
  }
  return body;
})

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM