mongoDB 查询 "WHERE _id >阈值" [英] mongoDB query "WHERE _id > threshold"

问题描述

我怎样才能有一个类似于 SQL "...WHERE _id > threshold" 的 mongo 查询

How can I have a mongo query similar to the SQL "...WHERE _id > threshold"

我尝试了以下方法，但没有给我任何结果.

I tried the following, but it doesn't give me any result.

 db.things.find(_id: {$gt: someid} });

是不是因为_id字段有点特殊，有格式?

Is it because the _id field is a little special, in that it has the format?

_id : {"$oid" : "someid"} 

推荐答案

比较喜欢和喜欢

mongo 中的 _id 键(默认情况下)不是字符串 - 它是 mongo objectId.

Compare like with like

The _id key in mongo is not (by default) a string - it is a mongo objectId.

您需要与相同类型进行比较才能获得有意义的结果:

You need to compare to the same type to get a meaningful result:

var ObjectId = require('mongodb').ObjectID;
var oid = new ObjectId();
db.things.find(_id: {$gt: oid});

不要阅读 mongoexport 文件

Mongo 导出文件如下所示:

Don't read mongoexport files

Mongo export files look like this:

{ "_id" : { "$oid" : "4f876b00c56da1fa6a000030" }, ...

这是一个对象 id 的 json 表示.Mongo 不希望您在实际查询数据库时使用这种语法.这将不起作用:

This is a json representation of an object id. Mongo doesn't want you to use that kind of syntax when actually querying the db. This will not work:

# will not work
db.things.find("_id.$oid": {$gt: "string"});

id 作为字符串

如果你有一个字符串形式的 id，你会这样做:

id as a string

If you have the id as a string, you'd do:

var ObjectId = require('mongodb').ObjectID;
var str = "123456789012345678901234";
var oid = new ObjectId(str);
db.things.find(_id: {$gt: oid});

id 作为部分字符串

如果您拥有的字符串不是有效的 oid(长度不是 24 个字符)，您只会从 mongo 中得到一个异常 - 或者根据您的驱动程序，一个新的 oid.如果您有部分对象 id，您可以用 0 填充以生成有效的 oid，因此允许通过部分对象 id 进行查找.例如:

id as a partial string

If the string you have is not a valid oid (not 24 chars long), you'll just get an exception from mongo - or depending on your driver, a new oid. If you have a partial object id you can pad with 0s to make a valid oid and therefore permit finding by partial object ids. e.g.:

var ObjectId = require('mongodb').ObjectID;
var oid = new ObjectId(str + "0000");
db.things.find(_id: {$gt: oid});

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