解析检索数据匹配几种不同的价值观

问题描述:

我在检索对象的解析,但它总是转到ParseException的解析检索数据匹配几种不同的价值观

这里是我的代码:

public void setItemListData(Context context, String className, 
      String[] value, String colName) { 
     ParseQuery query = new ParseQuery(className); 
     query.whereContainsAll("Store",Arrays.asList (value)); 
     query.whereEqualTo("Store","Asus"); 
     dataList = new ArrayList<String>(); 
     try { 
      List<ParseObject> dataHolder = query.find(); 
      for(int counter =0;counter<dataHolder.size();counter++){ 
       String datas = dataHolder.get(counter).getString(colName); 
       dataList.add(datas);  
      } 
     } catch (ParseException e) { 
      Toast.makeText(context,"No Internet Connection", 
      Toast.LENGTH_SHORT).show(); 
     } 
    } 

我不知道什么是错我的代码

在此先感谢

+0

当您使用'e.printStackTrace()'什么是例外? – Sam 2013-03-01 00:23:30

+0

等我没打印出来。我会检查并更新你 – user1708134 2013-03-01 00:24:18

+0

e.printStackTrace()= @all适用于数组字段 – user1708134 2013-03-01 00:33:43

Change query.whereContainsAll("Store",Arrays.asList (value)); 

query.whereContainedIn("Store",Arrays.asList (value)); 

查找多个查询:

private List<ParseObject> todos; 

代码找到多个值:

ParseQuery<ParseObject> query = new ParseQuery<ParseObject>("Your_Class_Name"); 
String[] names = {"Hiren", "Dhruv", "Pratik", "Miral", "Kirit"}; 
    query.whereContainedIn("Your_Column_Name_Means_Key", Arrays.asList(names)); 
    try { 
     todos = query.find(); 
     Log.i("No Of Record Found",todos.size()+""); 
     }catch (ParseException e) { 
     e.printStackTrace() 
     } 

你必须类名称你的列名称

完成