对所有人的相同错误响应因request.js而失败

Question

I'm using request.js for node.js.

I was wondering what would be the best way to return the same error for all failed responses.

Right now I'm returning the following json object for all fails:

res.json({success: false, message: 'An error has occurred'});

But I'm doing it per request, in each call. Like the following:

   request.post({
        uri: res.locals.baseUrl + 'myAction',
        qs: params
    }, function (error, response, body) {
        if (error || response.statusCode != 200) {
            res.json({success: false, message: 'An error has occurred'});
        }else{
            var data  = JSON.parse(body);
           res.json(data);
        }
    });

How can I deal with it in only one single place? Does request provides a way to do it? Or should I go for something like:

var failResponse ={success: false, message: 'An error has occurred'};

And then using this in every request:

res.json(failResponse);

Answer 1

Would something like this work for you?

function sendResp(error, response, body, callback) {
  if (error || response.statusCode != 200) {
    // Error occurred
    return callback(true, {
      success: false,
      message: 'An error has occurred'
    })
  }
  // No errors, just send the body
  callback(null, JSON.parse(body))
}

request.post({
  uri: res.locals.baseUrl + 'myAction',
  qs: params
}, function(error, response, body) {

  sendResp(error, response, body, function(error, msg){
     // If an error occured, return an error
    if(error) return res.json(msg)
    // Otherwise, display the response body
    res.json(msg)
  })
});

Or, going by your own suggestion (with a little clean up) you could do the following:

var failResponse = {
  success: false,
  message: 'An error has occurred'
};

request.post({
  uri: res.locals.baseUrl + 'myAction',
  qs: params
}, function(error, response, body) {
  // If there is an error, show it
  if (error || response.statusCode != 200) return res.json(failResponse);
  // No error, show response body
  res.json(JSON.parse(body));
});