从Java驱动程序传递多个$ Unwind对象 [英] Passing more than one $Unwind object from java driver

问题描述

mongo java驱动程序采用var args作为聚合方法，我有一个API，其中动态创建$unwind对象，并且其数量不固定.我如何通过Mongo Java驱动程序聚合方法传递它，因为它需要分别传递每个对象.我尝试通过将所有$unwind对象放入BasicDBList中并通过，但是失败.有人可以帮我解决问题吗?

The mongo java driver takes var args for aggregate method, I have an API in which $unwind objects get's created dynamically and its number is not fixed. how can I pass it through Mongo Java driver aggregate method, as it needs each object to be passed separately. I tried passing putting all the $unwind object in a BasicDBList and pass, but it fails. Can someone help me with some work around?

示例:

db.foo.aggregate({$unwind:items},{$unwind:item2})

，但是在运行时创建时，这些展开可能会有所不同.

, but these unwind may vary as it is getting created at runtime.

推荐答案

您无需创建BasicDBList.它是这样工作的:

you don't need to create a BasicDBList. This is how it works:

List<DBObject> unwindItems = new ArrayList<>();

if(<item2 is not null>){ //pseudo code
  DBObject unwindItem1 = new BasicDBObject("$unwind", "$item1");
  unwindItems.add(unwindItem1);
}
if(<item2 is not null>){ //pseudo code
  DBObject unwindItem2 = new BasicDBObject("$unwind", "$item2");
  unwindItems.add(unwindItem2);
}
//add any other dbObject in the list, it need not be an unwind operation, it could be match, project, group etc.

DBObject command = new BasicDBObject("aggregate", "foo");
command.put("pipeline", dbObjects);

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