如何获得超过1000个结果Parse [英] How to get more than 1000 results Parse

问题描述

我有一个云代码，有时返回超过1000个结果，我没有得到如何链接查询获得超过1000个结果在以下代码。

I have a cloud code which returns sometimes more than 1000 results , I am not getting how to chain query to get more than 1000 results in following code.

var User = Parse.Object.extend("Journal");
    var mainQuery = new Parse.Query(User);

    mainQuery.equalTo("user",username);     


    mainQuery.limit(1000);
    mainQuery.find({ 
        success: function(results) {
          count = 0;
          var avg = 0;

         console.log("self:"+results.length);
            for(j = 0; j < results.length ; j++)
            {
                var entry = results[j];

                if(strcmp1(day,"All") != 0)
                {
                    var dt = entry.get("created");

                    var weekDay = dt.getDay();
                    if(weekDay == day)
                    {
                       total += entry.get("Level");
                       count++;
                    }
                }
                else
                {
                    total += entry.get("Level");
                    count++;
                }
            }

            if(count > 1)
              avg = total / count;

            response.success(avg);  
        }
    });

limit函数用于提高默认限制100，每个查询结果返回周期最大为1000。

limit function is used for raising the default limit of 100 which is max 1000 per query result return cycle .

推荐答案

我目前正在使用此代码。这可以获取超过1000个结果。
这些值根据objectId进行排序。您可以根据createdAt通过相应地更改排序。
最终结果存储在结果数组中

I am using this code presently. This can fetch more than 1000 results. The values are sorted according to "objectId". You can sort according to "createdAt" by changing accordingly. Final results are stored in result array

      var result = [];
      var processCallback = function(res) {
                result = result.concat(res);
                if (res.length == 1000) {
                  process(res[res.length-1].id);
                  return;
                }   
                    //To print all the ids
                    for(var i=0;i<result.length;i++){
                        console.log(result[i].id);
                    }
                }

        var process = function(skip) {
            var query = new Parse.Query("ObjectName");

            if (skip) {
              query.greaterThan("objectId", skip);
            }
            query.limit(1000);
            query.ascending("objectId");
            query.find().then(function querySuccess(res) {
              processCallback(res);
            }, function queryFailed(error) {
                });
          }
      process(false);

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