Parse cloud code loop is only going through first instance

Question

I want the following loop to go through every instance of matchCenterItem, yet for some reason, it pings ebay using the properties of only the first instance. The console logs at the end of the function however, loop through all instances and log their respective properties.

  Parse.Cloud.define("MatchCenterTest", function(request, response) {

  var matchCenterItem = Parse.Object.extend("matchCenterItem");
  var query = new Parse.Query(matchCenterItem);

  var promises = [];

  query.limit(10);
  query.find().then(function(results) {
    for (i=0; i<results.length; i++) {

      url = 'http://svcs.ebay.com/services/search/FindingService/v1';
       promises.push(Parse.Cloud.httpRequest({
        url: url,
        params: {   
         'OPERATION-NAME' : 'findItemsByKeywords', 
         'SERVICE-VERSION' : '1.12.0',
         'SECURITY-APPNAME' : '*App ID goes here*',
         'GLOBAL-ID' : 'EBAY-US',
         'RESPONSE-DATA-FORMAT' : 'JSON',
         'REST-PAYLOAD&sortOrder' : 'BestMatch',
         'paginationInput.entriesPerPage' : '3',
         'outputSelector=AspectHistogram&itemFilter(0).name=Condition&itemFilter(0).value(0)' : results[i].get('itemCondition'),
         'itemFilter(1).name=MaxPrice&itemFilter(1).value' : results[i].get('maxPrice'),
         'itemFilter(1).paramName=Currency&itemFilter(1).paramValue' : 'USD',
         'itemFilter(2).name=MinPrice&itemFilter(2).value' : results[i].get('minPrice'),
         'itemFilter(2).paramName=Currency&itemFilter(2).paramValue' : 'USD',
         //'itemFilter(3).name=LocatedIn&itemFilter(3).Value' : request.params.itemLocation,
         'itemFilter(3).name=ListingType&itemFilter(3).value' : 'FixedPrice',
         'keywords' : results[i].get('searchTerm'),
        },
        // success: function (httpResponse) {
        //   // parses results
        //   var httpresponse = JSON.parse(httpResponse.text);

        //   response.success(httpresponse);
        //   console.log('MatchCenter Pinged eBay dude!');
        // },
        // error: function (httpResponse) {
        //   console.log('error!!!');
        //   response.error('Request failed with response code ' + httpResponse.status);
        // }
      }));

     console.log(results[i].get('itemCondition'));
     console.log(results[i].get('maxPrice'));
     console.log(results[i].get('minPrice'));
     console.log(results[i].get('searchTerm'));

    }
  });

Parse.Promise.when(promises).then(function(results) {

   var httpresponse = JSON.parse(httpResponse.text);
   response.success(httpresponse);
}, function(err) {
    console.log('error!!!');
});



});

Fosco · Accepted Answer · 2014-06-03 20:24:07Z

3

This is because the http request is asynchronous, and you're calling response.success in the completion handler for the first (and all) requests. Use the promise syntax and only complete when they are done. Simplified concept:

var promises = [];
for (...) {
   promises.push(Parse.Cloud.httpRequest({...}));  // no success/error params
}
Parse.Promise.when(promises).then(function(results) {
   response.success(...);
}, function(err) {

});

answered Jun 3, 2014 at 20:24

Fosco

38.6k7 gold badges90 silver badges101 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Ghobs Over a year ago

The logic makes sense, however, when adding

Parse.Promise.when(promises).then(function(results) {     var httpresponse = JSON.parse(httpResponse.text);    response.success(httpresponse); }, function(err) {     console.log('error!!!'); });

after the for loop, I'm getting an error stating that httpresponse isn't defined. Any idea why?

Ghobs Over a year ago

I've edited my code in the question to reflect how it now looks.

Fosco Over a year ago

You'll need to loop through the results variable, and access the response inside. I.e results[0].text

Collectives™ on Stack Overflow

Parse cloud code loop is only going through first instance

1 Answer 1

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related