获取表中的所有记录-Parse.com

我在一张表上有大约13000条记录（HashTag -classname）。我想在单个查询中检索它们。但解析只允许每个查询1000。任何其他方式获得的所有记录..获取表中的所有记录-Parse.com

ParseQuery<ParseObject> query = ParseQuery.getQuery("HashTag"); 
query.whereExists("Tag"); query.orderByAscending("Type"); query.setLimit(1000); 
query.findInBackground(new FindCallback<ParseObject>() { 

           @Override 
           public void done(List<ParseObject> list, 
             ParseException e) { 
            // TODO Auto-generated method stub 
            if (e == null) 
            { 
             if (list.size() > 0) { 
              for (int i = 0; i < list.size(); i++) { 
               ParseObject p = list.get(i); 
               String tagid = p.getString("Tag"); 
               String Type = p.getString("Type"); 
               class2 c2 = new class2(); 
               c2.type = "" + Type; 
               c2.tag = "" + tagid; 
               listClass2.add(c2); 



              } 

             }

来源

2014-09-23 Sree Reddy Menon

在分析数据浏览器中看到的 “更多” 下拉菜单，然后 “出口类” – 2014-09-24 18:04:05

//Declare a global variable for storing the complete data 
private static List<ParseObject>allObjects; 
allObjects=new ArrayList<ParseObject>(); 
ParseQuery<ParseObject>query3=ParseQuery.getQuery("HashTag"); 
query3.whereExists("Tag"); 
query3.setLimit(1000); 
query3.findInBackground(getallobjects()); 
int limit=1000; 
int skip=0; 
//callback method: 
private FindCallback<ParseObject>getallobjects(){ 
    return new FindCallback<ParseObject>(){ 
     @Override 
     public void done(List<ParseObject>list,ParseException e){ 
      allObjects.addAll(list); 
      if(list.size()==limit){ 
       skip=skip+limit; 
       ParseQuery<ParseObject>query=ParseQuery.getQuery("HashTag"); 
       query.setSkip(skip); 
       query.setLimit(limit); 
       query.findInBackground(getallobjects()); 
      }else{ 
       //you have full data in allobjects 
       for(int i=0;i<allObjects.size();i++){} 
      } 
     }}}

来源

2014-09-24 17:31:26

当然，你可以运行多个查询同一个表，用查询的skip财产1000每次递增：

获取总数通过query.count()记录，并用它来设置一个“跳过”变量
运行为每个1000只记录一个新的查询，相应地更新您跳过酒店
处理记录为正常时，每个查询返回

事情是这样的：

ParseQuery<ParseObject> query = ParseQuery.getQuery("HashTag"); 
    query.whereExists("Tag"); 
    query.countInBackground(new CountCallback() { 
     public void done(int count, ParseException e) { 
     if (e == null) { 
       // The count request succeeded. Run the query multiple times using the query count 
      int numQueries = Math.ceil(count/1000); //Gives you how many queries to run 
      for(int skipNum = 0; l < numQueries; l++){ 
      ParseQuery<ParseObject> query = ParseQuery.getQuery("HashTag"); 
      query.whereExists("Tag"); query.orderByAscending("Type"); 
      query.setLimit(skipNum * 1000); 
      query.findInBackground(new FindCallback<ParseObject>() { 
       //Run your query as normal here 

      } 
      } 
     } else { 
      // The request failed 
     } 
     }

来源

2014-09-23 15:20:58

这会增加API查询电话？ – 2014-09-23 15:42:40

是的，您必须为每个查询进行API调用，因此13000个记录= 13个查询。（顺便说一句，这也是解析自己所建议的解决方案 - 请参阅https://www.parse.com/questions/how-can-i-get-thousands-of-records-at-a- time） – 2014-09-24 13:01:17

ParseQuery<ParseObject> query = new ParseQuery<ParseObject>("TestObject"); 
      query.findInBackground(new FindCallback<ParseObject>() { 
       @Override 
       public void done(List<ParseObject> list, ParseException e) { 
        for(ParseObject p : list){ 

         Log.d("--", (String) p.get("foo")+p.getCreatedAt()); 
        } 
       } 
      });

来源

2015-07-01 00:04:30 vrbsm

获取表中的所有记录-Parse.com

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