如何在MongoDB中使用聚合框架排除_id而不包含其他字段？
首先让我们创建一个包含文档的集合-
> db.excludeIdDemo.insertOne({"StudentFirstName":"John","StudentAge":21});
{
"acknowledged" : true,
"insertedId" : ObjectId("5cd701a56d78f205348bc632")
}
> db.excludeIdDemo.insertOne({"StudentFirstName":"Robert","StudentAge":20});
{
"acknowledged" : true,
"insertedId" : ObjectId("5cd701af6d78f205348bc633")
}
> db.excludeIdDemo.insertOne({"StudentFirstName":"Chris","StudentAge":24});
{
"acknowledged" : true,
"insertedId" : ObjectId("5cd701b86d78f205348bc634")
}以下是在find()方法的帮助下显示集合中所有文档的查询-
> db.excludeIdDemo.find();
这将产生以下输出-
{ "_id" : ObjectId("5cd701a56d78f205348bc632"), "StudentFirstName" : "John", "StudentAge" : 21 }
{ "_id" : ObjectId("5cd701af6d78f205348bc633"), "StudentFirstName" : "Robert", "StudentAge" : 20 }
{ "_id" : ObjectId("5cd701b86d78f205348bc634"), "StudentFirstName" : "Chris", "StudentAge" : 24 }以下是使用聚合框架排除_id而不包含其他字段的查询-
> db.excludeIdDemo.aggregate(
{
$project :
{
_id : 0,
"StudentFirstName": 1
}
}
);这将产生以下输出-
{ "StudentFirstName" : "John" }
{ "StudentFirstName" : "Robert" }
{ "StudentFirstName" : "Chris" }热门推荐
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