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MongoTemplate:删除从find查找返回的元素列表

[英]MongoTemplate: remove list of elements returned from `find`

I am calling MongoTemplate.find(query, type) which returns a List<T> . 我正在调用MongoTemplate.find(query, type) ,它返回List<T> My question, how do I remove all elements in the list from the DB Collection in one DB call. 我的问题是,如何通过一个数据库调用从数据库集合中删除列表中的所有元素。 I know I could create a Query manually using id if I know the structure of T but what if I don't know what T is. 我知道如果知道T的结构,可以使用id手动创建Query ,但是如果我不知道T是什么,该怎么办。

MongoTemplate has a remove method that I can pass T . MongoTemplate有一个可以传递Tremove方法。 What I need is the equivalent removeAll(Collection<T> items, Class<T> type) 我需要的是等效的removeAll(Collection<T> items, Class<T> type)

EDIT per answer 每个答案编辑

I need to process each element and determine which ones to remove so I cannot simply pass the Query to remove . 我需要处理每个元素并确定要删除的元素,所以我不能简单地通过Query将其remove

Instead of calling find(Query, Class) and then remove all returned documents (a Collection or one-by-one), you can simply call remove(Query, Class) . 无需调用find(Query,Class)然后删除所有返回的文档(Collection或一对一),只需调用remove(Query,Class)即可 It's much better that way since you don't have to return all documents to the client, but issue a remove query in the DB itself. 这种方法要好得多,因为您不必将所有文档都返回给客户端,而是可以在数据库本身中发出删除查询。

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