Consider the following Java Class.
class Transaction {
#Id
public String id;
public String firstProperty;
public String secondProperty;
}
In Java following code is executed :
(mongoTemplate is of type Mongoperations)
Transaction transaction = new Transaction("T1");
transaction.setFirstProperty("first");
Query query = Query.query(Criteria.where("_id").is("T1"));
DBObject dbObject = new BasicDBObject();
mongoTemplate.getConverter().write(transactionInfo, dbObject);
Update update = Update.fromDBObject(dbObject);
mongoTemplate.upsert(query , update , TransactionInfo.class);
Following document is created.
{
_id:123,
firstProperty: "first"
}
If this piece of code is executed later :
Transaction transaction = new Transaction("T1");
transaction.setSecondProperty("second");
Query query = Query.query(Criteria.where("_id").is("T1"));
mongoTemplate.upsert(query , update , TransactionInfo.class);
Expected Document :
{
_id:123,
firstProperty: "first",
secondProperty: "second"
}
Actual Document:
{
_id:123,
secondProperty: "second"
}
From what I read in MongoDB docs I expect the document to be updated with "secondProperty" but it results in the removal of "firstProperty" . I think the document is getting created again, instead of getting updated. Please let me know if I am missing something.
Related
I am trying to sort my table's content on the backend side, so I am sending org.springframework.data.domain.Pageable object to controller. It arrives correctly, but at the repository I am getting org.hibernate.hql.internal.ast.InvalidPathException. Somehow the field name I would use for sorting gets an org. package name infront of the filed name.
The Pageable object logged in the controller:
Page request [number: 0, size 10, sort: referenzNumber: DESC]
Exception in repository:
Invalid path: 'org.referenzNumber'","logger_name":"org.hibernate.hql.internal.ast.ErrorTracker","thread_name":"http-nio-8080-exec-2","level":"ERROR","level_value":40000,"stack_trace":"org.hibernate.hql.internal.ast.InvalidPathException: Invalid path: 'org.referenzNumber'\n\tat org.hibernate.hql.internal.ast.util.LiteralProcessor.lookupConstant(LiteralProcessor.java:111)
My controller endpoint:
#GetMapping(value = "/get-orders", params = { "page", "size" }, produces = { MediaType.APPLICATION_JSON_VALUE })
public ResponseEntity<PagedModel<KryptoOrder>> getOrders(
#ApiParam(name = "searchrequest", required = true) #Validated final OrderSearchRequest orderSearchRequest,
#PageableDefault(size = 500) final Pageable pageable, final BindingResult bindingResult,
final PagedResourcesAssembler<OrderVo> pagedResourcesAssembler) {
if (bindingResult.hasErrors()) {
return ResponseEntity.badRequest().build();
}
PagedModel<Order> orderPage = PagedModel.empty();
try {
var orderVoPage = orderPort.processOrderSearch(resourceMapper.toOrderSearchRequestVo(orderSearchRequest), pageable);
orderPage = pagedResourcesAssembler.toModel(orderVoPage, orderAssembler);
} catch (MissingRequiredField m) {
log.warn(RESPONSE_MISSING_REQUIRED_FIELD, m);
return ResponseEntity.badRequest().build();
}
return ResponseEntity.ok(orderPage);
}
the repository:
#Repository
public interface OrderRepository extends JpaRepository<Order, UUID> {
static final String SEARCH_ORDER = "SELECT o" //
+ " FROM Order o " //
+ " WHERE (cast(:partnerernumber as org.hibernate.type.IntegerType) is null or o.tradeBasis.account.retailpartner.partnerbank.partnerernumber = :partnerernumber)"
+ " and (cast(:accountnumber as org.hibernate.type.BigDecimalType) is null or o.tradeBasis.account.accountnumber = :accountnumber)"
+ " and (cast(:orderReference as org.hibernate.type.LongType) is null or o.tradeBasis.referenceNumber = :orderReference)"
+ " and (cast(:orderReferenceExtern as org.hibernate.type.StringType) is null or o.tradeBasis.kundenreferenceExternesFrontend = :orderReferenceExtern)"
+ " and (cast(:dateFrom as org.hibernate.type.DateType) is null or o.tradeBasis.timestamp > :dateFrom) "
+ " and (cast(:dateTo as org.hibernate.type.DateType) is null or o.tradeBasis.timestamp < :dateTo) ";
#Query(SEARCH_ORDER)
Page<Order> searchOrder(#Param("partnerernumber") Integer partnerernumber,
#Param("accountnumber") BigDecimal accountnumber, #Param("orderReference") Long orderReference,
#Param("orderReferenceExtern") String orderReferenceExtern, #Param("dateFrom") LocalDateTime dateFrom,
#Param("dateTo") LocalDateTime dateTo, Pageable pageable);
}
Update:
I removed the parameters from the sql query, and put them back one by one to see where it goes sideways. It seems as soon as the dates are involved the wierd "org." appears too.
Update 2:
If I change cast(:dateTo as org.hibernate.type.DateType) to cast(:dateFrom as date) then it appends the filed name with date. instead of org..
Thanks in advance for the help
My guess is, Spring Data is confused by the query you are using and can't properly append the order by clause to it. I would recommend you to use a Specification instead for your various filters. That will not only improve the performance of your queries because the database can better optimize queries, but will also make use of the JPA Criteria API behind the scenes, which requires no work from Spring Data to apply an order by specification.
Since your entity Order is named as the order by clause of HQL/SQL, my guess is that Spring Data tries to do something stupid with the string to determine the alias of the root entity.
I want to get one Embedded Document with a specific field (version) from an array with mongodb and spring boot.
This is the data structure:
{
"_id": 5f25882d28e40663719d0b52,
"versions": [
{
"versionNr": 1
"content": "This is the first Version of some Text"
},
{
"versionNr": 2
"content": "This is the second Version of some Text"
},
...
]
...
}
Here are my entities:
#Data
#Document(collection = "letters")
public class Letter {
#Id
#Field("_id")
private ObjectId _id;
#Field("versions")
private List<Version> versions;
}
//There is no id for embedded documents
#Data
#Document(collection = "Version")
public class Version{
#Field("content")
private String content;
#Field("version")
private Long version;
}
And this is the query that doesn't work. I think the "join" isn't correct. But can't figure out the right way.
public Optional<Version> findByIdAndVersion(ObjectId id, Long version) {
Query query = new Query(Criteria.where("_id").is(id).and("versions.version").is(version));
return Optional.ofNullable(mongoTemplate.findOne(query,Version.class,"letters"));
}
}
EDIT: This is a working Aggregation, I'm sure it isn't a pretty solution but it works
#Override
public Optional<Version> findByIdAndVersion(ObjectId id, Long version) {
MatchOperation match = new MatchOperation(Criteria.where("_id").is(id).and("versions.version").is(version));
Aggregation aggregate = Aggregation.newAggregation(
match,
Aggregation.unwind("versions"),
match,
Aggregation.project()
.andInclude("versions.content")
.andInclude("versions.version")
);
AggregationResults<Version> aggregateResult = mongoTemplate.aggregate(aggregate, "letters", Version.class);
Version version = aggregateResult.getUniqueMappedResult();
return Optional.ofNullable(mongoRawPage);
}
Query query = new Query(Criteria.where("_id").is(id).and("versions.version").is(version));
return Optional.ofNullable(mongoTemplate.findOne(query,Version.class,"letters"));
You are querying the Letter document but your entity class is specified as Version.class, since findOne from MongoDB doesn't return the subdocument by itself but rather the whole document, you need to have Letter.class as return type and filter (project) what fields to get back. So you can either project the single version subdocument that you want to receive, like so:
Query query = new Query()
.addCriteria(Criteria.where("_id").is(id).and("versions.version").is(version))
.fields().position("versions", 1);
Optional.ofNullable(mongoTemplate.findOne(query, Letter.class))
.map(Letter::getVersions)
.findFirst()
.orElse(null);
or use aggregation pipeline:
newAggregation(
Letter.class,
match(Criteria.where("_id").is(id)),
unwind("versions"),
replaceRoot("versions"),
match(Criteria.where("version").is(version))),
Version.class)
Note -- I typed this on a fly.
I am new to mongodb and struggling to understand how document update works.
I have a document called 'menu':
{
"someId":"id123",
"someProperty":"property123",
"list" : [{
"innerProperty":"property423"
}]
}
which maps to my entity:
#Document(collection = "menu")
public class Menu {
#Id
private String id;
private String someid;
private String someProperty;
private List<SomeClass> list;
// accessors
}
when I try to find and update this document like this it does not update the document. It sure does find the menu as as it returns the original entity with Id:
#Override
public Menu update(Menu menu) {
Query query = new Query(
Criteria.where("someId").is(menu.getSomeId()));
Update update = Update.update("menu", menu);
return mongoOperations.findAndModify(query, update,
FindAndModifyOptions.options().returnNew(true), Menu.class);
}
But if I change it to this, it works:
#Override
public Menu update(Menu menu) {
Query query = new Query(
Criteria.where("someId").is(menu.getSomeId()));
Update update = new Update().set("someProperty", menu.getSomeProperty())
.set("list", menu.getList());
return mongoOperations.findAndModify(query, update,
FindAndModifyOptions.options().returnNew(true), Menu.class);
}
I don't really like this second method where each element of the document is individually set, as you might imagine I have a rather large document and is prone to errors.
Why does the first method not work? And what could be a better approach to update the document?
Check out the docs for findAndModify - it returns the version of the document before the fields were modified. If you do a new find() straight after, you will see that your changes were actually saved to MongoDB.
I am having trouble to query that, I want to find all the docs that contains the id "5418a26ce4b0e4a40ea1d548" in the individualUsers field. would be esp useful if you know how to do that in Spring Data MongoDB query.
db.collection.find({individualUser:{"5418a26ce4b0e4a40ea1d548"}})
example of one doc
{ "_id" : ObjectId("5418c3b9e4b03feec4345602"), "creatorId" : "5418a214e4b0e4a40ea1d546", "individualUsers" : { "5418a26ce4b0e4a40ea1d548" : null, "5418a278e4b0e4a40ea1d54a" : null } }
Update #001
Entity code
#Document
class Idea{
#Id
String id;
String creatorId;
Map<String,String> individualUsers;
/*getter and setter omitted*/
}
Interface
public interface IdeaRepository extends MongoRepository<Idea,String> {
}
Update #002
So when spring-mongodb saves hashmap to json, it will look like
"individualUsers" : { "5418a26ce4b0e4a40ea1d548" : null, "5418a278e4b0e4a40ea1d54a" : null }
In java program I can easily get the data using the key value. but in the mongodb query, I can't query the key?
so the question is can I query inside the "individualUsers": {} with the key ??
I have this request working good with queryDSL :
Iterable<AO> query_result = new JPAQuery(entityManager).from(ao)
.leftJoin( ao.lots , lot )
.leftJoin( ao.acs , ac )
.where(where).distinct()
.list(ao);
But what is its equivalent if we use it with spring data jpa
ao_respository.findAll(Predicate arg0, Pageable arg1);
Because i want to return a Page and just with querydsl it doesn't implement Page without spring data jpa.
I try to put my where in Predicate arg0 but i got this exception
Undeclared path 'lot '. Add this path as a source to the query to be able to reference it
where lot is declared as QLot lot = QLot.lot;
I created my own Page class and executed the query like this:
JPAQuery query = new JPAQuery(entityManager).from(ao)
.leftJoin( .. ).fetch()
.leftJoin( .. ).fetch()
...
.where(where)
MaPage<AO> page = new MaPage<AO>();
page.number = pageNumber+1;
page.content = query.offset(pageNumber*pageSize).limit(pageSize).list(ao);
page.totalResult = query.count();
My Page class:
public class MaPage<T> {
public List<T> content;
public int number;
public Long totalResult;
public Long totalPages;
...
}
It works but I got this warning
nov. 21, 2014 6:48:54 AM
org.hibernate.hql.internal.ast.QueryTranslatorImpl list WARN:
HHH000104: firstResult/maxResults specified with collection fetch;
applying in memory!
Returning a Page:
JPAQuery query =
...
.orderBy(getOrderSpecifiers(pageable, MyEntity.class))
.limit(pageable.getPageSize())
.offset(pageable.getOffset());
long total = query.fetchCount();
List<MyEntity> content = query.fetch();
return new PageImpl<>(content, pageable, total);
And I created this function to get OrderSpecifier:
private OrderSpecifier[] getOrderSpecifiers(#NotNull Pageable pageable, #NotNull Class klass) {
// orderVariable must match the variable of FROM
String className = klass.getSimpleName();
final String orderVariable = String.valueOf(Character.toLowerCase(className.charAt(0))).concat(className.substring(1));
return pageable.getSort().stream()
.map(order -> new OrderSpecifier(
Order.valueOf(order.getDirection().toString()),
new PathBuilder(klass, orderVariable).get(order.getProperty()))
)
.toArray(OrderSpecifier[]::new);
}
If you have a working, complex query in querydsl and you want to use springdata pagination, you have to:
make your querydsl/repository method return Page<T>
Page<YourEntity> yourSelect(Pageable aPageable)
use querydsl offset and limit to page your result set
List<YourEntity> theResultList = jpaQueryFactory
.select(<whatever complext jpaquery you like>)
.offset(aPageable.getOffset())
.limit(aPageable.getPageSize())
.fetch();
provide a LongSuplier counting all available results with respect to your query and use PageableExecutionUtils to return the result as Page
final long theCount = jpaQueryFactory
.selectFrom(<your select to count all results>)
.fetchCount();
return PageableExecutionUtils.getPage(theResultList, aPageable, () -> theCount);