Alw*_*ing 16 spring mongodb spring-data aggregation-framework spring-data-mongodb
如何将以下 MongoDB 查询转换为 Java Spring 应用程序要使用的查询?我找不到使用pipeline提供的查找方法的方法。
这是我试图转换的查询。我还想指出,我没有使用,$unwind因为我希望将deliveryZipCodeTimings保留为返回对象中的分组集合。
db.getCollection('fulfillmentChannel').aggregate([
{
$match: {
"dayOfWeek": "SOME_VARIABLE_STRING_1"
}
},
{
$lookup: {
from: "deliveryZipCodeTiming",
let: { location_id: "$fulfillmentLocationId" },
pipeline: [{
$match: {
$expr: {
$and: [
{$eq: ["$fulfillmentLocationId", "$$location_id"]},
{$eq: ["$zipCode", "SOME_VARIABLE_STRING_2"]}
]
}
}
},
{
$project: { _id: 0, zipCode: 1, cutoffTime: 1 }
}],
as: "deliveryZipCodeTimings"
}
},
{
$match: {
"deliveryZipCodeTimings": {$ne: []}
}
}
])
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Alw*_*ing 13
基于@dnickless 提供的信息,我能够解决这个问题。我将发布完整的解决方案,希望它在未来对其他人有所帮助。
我正在使用mongodb-driver:3.6.4
首先,我必须创建一个自定义聚合操作类,以便我可以传入一个自定义 JSON mongodb 查询以用于聚合操作。这将允许我pipeline在我使用$lookup的驱动程序版本不支持的范围内使用。
public class CustomProjectAggregationOperation implements AggregationOperation {
private String jsonOperation;
public CustomProjectAggregationOperation(String jsonOperation) {
this.jsonOperation = jsonOperation;
}
@Override
public Document toDocument(AggregationOperationContext aggregationOperationContext) {
return aggregationOperationContext.getMappedObject(Document.parse(jsonOperation));
}
}
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现在我们能够将自定义 JSON 查询传递到我们的 mongodb spring 实现中,剩下的就是将这些值插入到TypedAggregation查询中。
public List<FulfillmentChannel> getFulfillmentChannels(
String SOME_VARIABLE_STRING_1,
String SOME_VARIABLE_STRING_2) {
AggregationOperation match = Aggregation.match(
Criteria.where("dayOfWeek").is(SOME_VARIABLE_STRING_1));
AggregationOperation match2 = Aggregation.match(
Criteria.where("deliveryZipCodeTimings").ne(Collections.EMPTY_LIST));
String query =
"{ $lookup: { " +
"from: 'deliveryZipCodeTiming'," +
"let: { location_id: '$fulfillmentLocationId' }," +
"pipeline: [{" +
"$match: {$expr: {$and: [" +
"{ $eq: ['$fulfillmentLocationId', '$$location_id']}," +
"{ $eq: ['$zipCode', '" + SOME_VARIABLE_STRING_2 + "']}]}}}," +
"{ $project: { _id: 0, zipCode: 1, cutoffTime: 1 } }]," +
"as: 'deliveryZipCodeTimings'}}";
TypedAggregation<FulfillmentChannel> aggregation = Aggregation.newAggregation(
FulfillmentChannel.class,
match,
new CustomProjectAggregationOperation(query),
match2
);
AggregationResults<FulfillmentChannel> results =
mongoTemplate.aggregate(aggregation, FulfillmentChannel.class);
return results.getMappedResults();
}
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我想添加这个我的解决方案,它在某些方面重复之前发布的解决方案。
对于 Mongo 驱动程序 v3.x 我得出以下解决方案:
import java.util.Collections;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
import com.mongodb.BasicDBList;
import com.mongodb.BasicDBObject;
import com.mongodb.util.JSON;
import org.bson.Document;
import org.springframework.data.mongodb.core.aggregation.AggregationOperation;
import org.springframework.data.mongodb.core.aggregation.AggregationOperationContext;
public class JsonOperation implements AggregationOperation {
private List<Document> documents;
public JsonOperation(String json) {
Object root = JSON.parse(json);
documents = root instanceof BasicDBObject
? Collections.singletonList(new Document(((BasicDBObject) root).toMap()))
: ((BasicDBList) root).stream().map(item -> new Document((Map<String, Object>) ((BasicDBObject) item).toMap())).collect(Collectors.toList());
}
@Override
public Document toDocument(AggregationOperationContext context) {
// Not necessary to return anything as we override toPipelineStages():
return null;
}
@Override
public List<Document> toPipelineStages(AggregationOperationContext context) {
return documents;
}
}
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然后假设在某些资源中给出了聚合步骤aggregations.json:
[
{
$match: {
"userId": "..."
}
},
{
$lookup: {
let: {
...
},
from: "another_collection",
pipeline: [
...
],
as: "things"
}
},
{
$sort: {
"date": 1
}
}
]
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可以按如下方式使用上面的类:
import static org.springframework.data.mongodb.core.aggregation.Aggregation.newAggregation;
Collection<ResultDao> results = mongoTemplate.aggregate(newAggregation(new JsonOperation(resourceToString("aggregations.json", StandardCharsets.UTF_8))), "some_collection", ResultDao.class).getMappedResults();
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由于JSON类已从 Mongo v4 中删除,我将类重写如下:
import java.util.Collections;
import java.util.List;
import org.bson.Document;
import org.springframework.data.mongodb.core.aggregation.AggregationOperation;
import org.springframework.data.mongodb.core.aggregation.AggregationOperationContext;
public class JsonOperation implements AggregationOperation {
private List<Document> documents;
private static final String DUMMY_KEY = "dummy";
public JsonOperation(String json) {
documents = parseJson(json);
}
static final List<Document> parseJson(String json) {
return (json.startsWith("["))
? Document.parse("{\"" + DUMMY_KEY + "\": " + json + "}").getList(DUMMY_KEY, Document.class)
: Collections.singletonList(Document.parse(json));
}
@Override
public Document toDocument(AggregationOperationContext context) {
// Not necessary to return anything as we override toPipelineStages():
return null;
}
@Override
public List<Document> toPipelineStages(AggregationOperationContext context) {
return documents;
}
@Override
public String getOperator() {
return documents.iterator().next().keySet().iterator().next();
}
}
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但由于字符串操作,现在的实现有点难看。如果有人对如何以更优雅的方式解析对象数组有更好的想法,请编辑这篇文章或发表评论。理想情况下,Mongo 核心中应该有一些方法允许解析 JSON 对象或列表(返回BasicDBObject/BasicDBList或Document/ List<Document>)。
Document另请注意,我跳过了在方法中转换实例的步骤toPipelineStages(),因为在我的情况下没有必要:
@Override
public List<Document> toPipelineStages(AggregationOperationContext context) {
return documents.stream().map(document -> context.getMappedObject(document)).collect(Collectors.toList());
}
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