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java - MongoTemplate聚合返回 "IllegalArgumentException: Invalid reference"

转载 作者:行者123 更新时间:2023-12-01 17:27:27 25 4
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给定一个包含以下文档的集合:

{
"_id": {
"$oid": "5e8f83289d3a48132e9b6e34"
},
"resource-id": "my-default-network-map",
"uri": "http://alto.example.com/networkmap",
"mappings": [
{
"version-tag": "X",
"address-aggregations": [{
"pid": "PID1",
"ipv4-addresses": ["192.0.2.0/24", "198.51.100.0/25"]
}, {
"pid": "PID2",
"ipv4-addresses": ["198.51.100.128/25"]
}, {
"pid": "PID3",
"ipv4-addresses": ["0.0.0.0/0"],
"ipv6-addresses": ["::/0"]
}]
},
{
"version-tag": "Y",
"address-aggregations": [{
"pid": "PID1",
"ipv4-addresses": ["192.0.2.0/24", "198.51.100.0/25"]
}, {
"pid": "PID2",
"ipv4-addresses": ["198.51.100.128/25"]
}]
}],
"_class": "com.example.restservice.entity.NetworkMapEntity"
}

并且假设更多这些文档与其他“resource-id”值一起存在,我希望返回我之前输入的相同文档,但“mappings”数组仅包含 version-tag = X 的对象和“address-aggregations”数组仅包含[pidx,pidy,...]中带有“pid”的对象。我目前能想到的最好的方法是以下聚合管道:

[
{
'$match': {
'resource-id': 'my-default-network-map'
}
}, {
'$unwind': {
'path': '$mappings'
}
}, {
'$match': {
'mappings.version-tag': 'X'
}
}, {
'$unwind': {
'path': '$mappings.address-aggregations'
}
}, {
'$match': {
'mappings.address-aggregations.pid': {
'$in': [
'PID1', 'PID2'
]
}
}
}, {
'$group': {
'_id': '$_id',
'resource-id': {
'$first': '$resource-id'
},
'uri': {
'$first': '$uri'
},
'version-tag': {
'$first': '$mappings.version-tag'
},
'address-aggregations': {
'$push': '$mappings.address-aggregations'
}
}
}, {
'$project': {
'_id': 1,
'resource-id': 1,
'uri': 1,
'mappings.version-tag': '$version-tag',
'mappings.address-aggregations': '$address-aggregations'
}
}, {
'$group': {
'_id': '$_id',
'resource-id': {
'$first': '$resource-id'
},
'uri': {
'$first': '$uri'
},
'mappings': {
'$push': '$mappings'
}
}
}
]

这在 mongodb 客户端 shell 上返回了预期结果。但是,当我将此类对象转换为带有聚合对象的 MongoTemplate 查询时:

Aggregation aggregation = newAggregation(
match(Criteria.where("resource-id").is(resourceId)),
unwind("mappings"),
match(Criteria.where("mappings.version-tag").is(versionTag)),
unwind("mappings.address-aggregations"),
match(Criteria.where("mappings.address-aggregations.pid").in(pids)),
group("$_id")
.first("resource-id").as("resource-id")
.first("uri").as("uri")
.first("mappings.version-tag").as("version-tag")
.push("mappings.address-aggregations").as("address-aggregations"),
project("_id", "resource-id", "uri")
.and("version-tag").as("mappings.version-tag")
.and("address-aggregations").as("mappings.address-aggregations"),
group("_id")
.first("resource-id").as("resource-id")
.first("uri").as("uri")
.push("mappings").as("mappings")
);


AggregationResults<NetworkMapEntity> aggregationResults = mongoTemplate.aggregate(aggregation, "NetworkMaps", NetworkMapEntity.class);

并测试它,抛出以下异常:java.lang.IllegalArgumentException:无效的引用'mappings'!

搜索周围我发现有类似问题的人说这是 springframework.data 依赖项的错误,但在一年前的版本上,所以我认为我对聚合对象的构造可能是问题所在。我在解决方案的 Java 代码部分做错了什么吗?

最佳答案

虽然我仍然不明白为什么当前的解决方案不起作用,并且想了解为什么不起作用,但我按如下方式更改了查询,现在它似乎按预期工作:

Aggregation aggregation = newAggregation(
match(Criteria.where("resource-id").is(resourceId)),
unwind("mappings"),
match(Criteria.where("mappings.version-tag").is(versionTag)),
unwind("mappings.address-aggregations"),
match(Criteria.where("mappings.address-aggregations.pid").in(pids)),
group("_id", "resource-id", "uri")
.first("mappings.version-tag").as("version-tag")
.push("mappings.address-aggregations").as("address-aggregations")
group("_id._id")
.first("_id.resource-id").as("resource-id")
.push(
new BasicDBObject("version-tag", "$version-tag")
.append("address-aggregations", "$address-aggregations")).as("mappings")
);

关于java - MongoTemplate聚合返回 "IllegalArgumentException: Invalid reference",我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/61190212/

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