简体   繁体   English

mongodb条件查询大于$ gt返回零结果

[英]mongodb conditional query greater than $gt returns zero results

I have a mongodb database which has a users collection containing the following document 我有一个mongodb数据库,其中包含一个包含以下文档的用户集合

{
  "_id": ObjectId("5161446e03642eab4a818fcd"),
  "id": "35",
  "userInfo": {
  "name": "xyz",
  "rollNumber": 121
   }
}

I want to get all the rows whose id is greater than a specific value 我想获取所有ID大于特定值的行

@GET
@Path("/query")
@Produces({ MediaType.APPLICATION_JSON })
 public List<String> getLactionInfo(@QueryParam("from") int from) {
    BasicDBObject query = new BasicDBObject();
    // check if ids greater than 0
    query.put("id", new BasicDBObject("$gt", from));
    // get the collection of users
    DBCursor cursor = getTable("users").find(query);
    List<String> listUsers = new ArrayList<String>();
    while (cursor.hasNext()) {
        DBObject object = cursor.next();
        String id = cursor.next().get("_id").toString();
        object.put("_id", id);
        String objectToString = object.toString();
        listUsers.add(objectToString);

    }

    return listUsers;
}

When I debugged my code its showing that the listUsers is null.Also when I manually run the following query in the console I get no results. 当我调试代码时,它显示listUsers为null。此外,当我在控制台中手动运行以下查询时,我也没有结果。

db.users.find({id:{$gt:60}})

The id in your sample data is stored as a string . 样本数据中的idstring形式存储。 So, the $gt check is trying to compare an integer to a string . 因此, $gt检查试图将integerstring进行比较。

If you switch your id value to an integer , it should work as expected. 如果将id值切换为integer ,则它将按预期工作。

For example: 例如:

db.test.insert( { "id": 40, "name": "wired" } )
db.test.insert( { "id": 60, "name": "prairie" } )
db.test.insert( { "id": 70, "name": "stack" } )
db.test.insert( { "id": 80, "name": "overflow" } )
db.test.insert( { "id": "90", "name": "missing" } )

Then, the test: 然后,测试:

> db.test.find({"id": { "$gt": 60 }}).pretty()
{
        "_id" : ObjectId("516abfdf8e7f7f35107081cc"),
        "id" : 70,
        "name" : "stack"
}
{
        "_id" : ObjectId("516abfe08e7f7f35107081cd"),
        "id" : 80,
        "name" : "overflow"
}

For a quick data fix up you could do something like this from the shell (paste as one line and change myCollection to your collection name: 为了快速修复数据,您可以在shell中执行以下操作(将其粘贴为一行,并将myCollection更改为您的集合名称:

db.myCollection.find().forEach(function(doc) { doc.id = parseInt(doc.id, 10); 
     db.test.save(doc); })

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM