Mongo db java驱动程序查询转换
我有以下数据结构Mongo db java驱动程序查询转换,java,mongodb,aggregation-framework,mongodb-java,mongodb-java-3.3.0,Java,Mongodb,Aggregation Framework,Mongodb Java,Mongodb Java 3.3.0,我有以下数据结构 [{ "id": "1c7bbebd-bc3d-4352-9ac0-98c01d13189d", "version": 0, "groups": [ { "internalName": "Admin group", "fields": [ { "internalName": "Is verified",
[{
"id": "1c7bbebd-bc3d-4352-9ac0-98c01d13189d",
"version": 0,
"groups": [
{
"internalName": "Admin group",
"fields": [
{
"internalName": "Is verified",
"uiProperties": {
"isShow": true
}
},
{
"internalName": "Hide",
"uiProperties": {
"isHide": false
}
},
...
]
},
...
]
},
{
"id": "2b7bbebd-bc3d-4352-9ac0-98c01d13189d",
"version": 0,
"groups": [
{
"internalName": "User group",
"fields": [
{
"internalName": "Is verified",
"uiProperties": {
"isShow": true
}
},
{
"internalName": "Blocked",
"uiProperties": {
"isBlocked": true
}
},
...
]
},
...
]
},
...
]
字段的内部名称可以重复。我想按group.field.internalName分组并剪切数组(用于分页),然后获得如下输出:
{
"totalCount": 3,
"items": [
{
"internalName": "Blocked"
},
{
"internalName": "Hide"
},
{
"internalName": "Is verified"
}
]}
我写了一个有效的查询
db.layouts.aggregate(
{
$unwind : "$groups"
},
{
$unwind : "$groups.fields"
},
{
$group: {
"_id" : {
"internalName" : "$groups.fields.internalName",
},
"internalName" : {
$first : "$groups.fields.internalName"
}
}
},
{
$group: {
"_id" : null,
"items" : {
$push : "$$ROOT"
},
"totalCount" : {
$sum : 1
}
}
},
{
$project: {
"items" : {
$slice : [ "$items", 0, 20 ]
},
"totalCount": 1
}
})
但是我有一个问题,就是如何将它转换成JavaAPI。请注意,我需要使用mongoTemplate方法。这是我所拥有的,也是我震惊的地方
final List<AggregationOperation> aggregationOperations = new ArrayList<>();
aggregationOperations.add(unwind("groups"));
aggregationOperations.add(unwind("groups.fields"));
aggregationOperations.add(
group("groups.fields.internalName")
.first("groups.fields.internalName").as("internalName")
);
aggregationOperations.add(
group()
.push("$$ROOT").as("fields")
.sum("1").as("totalCount") // ERROR only string ref can be placed, but i need a number?
);
aggregationOperations.add(
project()
.andInclude("totalCount")
.and("fields").slice(size, page * size)
);
final Aggregation aggregation = newAggregation(aggregationOperations);
mongoTemplate.aggregate(aggregation, LAYOUTS, FieldLites.class).getMappedResults()
final List aggregationOperations=new ArrayList();
aggregationOperations.add(展开(“组”);
aggregationOperations.add(展开(“groups.fields”);
aggregationOperations.add(
组(“组.字段.内部名称”)
.first(“groups.fields.internalName”).as(“internalName”)
);
aggregationOperations.add(
组()
.push(“$$ROOT”).as(“字段”)
.sum(“1”).as(“totalCount”)//只能放置字符串ref,但我需要一个数字吗?
);
aggregationOperations.add(
项目()
.和包括(“总数”)
.和(“字段”).切片(大小、页面*大小)
);
最终聚合=新聚合(聚合操作);
聚合(聚合、布局、FieldLites.class).getMappedResults()
在这个查询中,sum()有问题,因为我只能按api放置一个字符串ref(但需要一个数字),而在项目操作中,我得到了一个异常
java.lang.IllegalArgumentException:引用“totalCount”无效!]根本原因
你能帮我翻译这个查询吗?你可以使用
count
group()
.push("$$ROOT").as("fields")
.count().as("totalCount")
您可以使用
count
group()
.push("$$ROOT").as("fields")
.count().as("totalCount")