Order Page by different Date Properties with Spring Data - java

I am using Spring Data to access a SQL Database with mapped objects.
I have got an object which has multiple date properties which are relevant in different states of the object.
public class Foo {
Date deadlineOne; // relevant in State.ONE
Date deadlineTwo; // relevant in State.TWO
Date deadlineThree; // relevant in State.THREE
State state; // implemented via persisted Enum
}
Now, in the UI I want to display these objects with a property called something like 'Next Deadline' and also be able to sort by this aspect. This 'Next Deadline' should always be the earliest Date relevant by the related state. So for example, if I have the following objects:
O1{state:ONE, deadlineOne: '2016-06-14', deadlineTwo: '2016-06-16', deadlineThree: '2016-06-19'}
O2{state:TWO, deadlineOne: '2016-06-12', deadlineTwo: '2016-06-13', deadlineThree: '2016-06-20'}
O3{state:THREE, deadlineOne: '2016-06-15', deadlineTwo: '2016-06-18', deadlineThree: '2016-06-21'}
I want them to be displayed like this when sorted by 'Next Deadline'
Object | Next Deadline | State
O2 | 2016-06-13 | TWO
O1 | 2016-06-14 | ONE
O3 | 2016-06-21 | THREE
I just can not figure out how to implement this functionality with Spring Data. Up to this point I have been using a PageRequest and got basic sorting working by just passing a property of my object. But that way I can not mix the sorting-properties the way I need to.
What I've tried so far
Implementing this functionality via JPA Query Annotaion using SQL Case syntax like this in a JpaRepository interface:
#Query("SELECT f FROM Foo f ORDER BY CASE
WHEN f.state = State.ONE THEN f.deadlineOne
WHEN f.state = State.TWO THEN f.deadlineTwo
WHEN f.state = State.THREE THEN f.deadlineThree")
Page<Foo> findSortedByNextDeadline(Pageable pageable);
The interpreter just cant seem to accept this and gives no useful error information.
Question
Is there any way to make this work with Spring Data and the right Query/PageRequest or do I have to maintain a seperate nextDeadline property or something of that matter?

Alright I got it working, and here is how:
Like Dimpre Jean-Sébastien pointed out in his comment I was missing the END keyword. I also had to add an explicit #Enumerated(EnumType.ORDINAL) to my state-property (which is only relevant in my particular case, its should work with String types just aswell). Also I referenced the Enum for my state with its full package name. In the end it looked something like this:
#Query("SELECT f FROM Foo f ORDER BY CASE
WHEN f.state = com.package.State.ONE THEN f.deadlineOne
WHEN f.state = com.package.State.TWO THEN f.deadlineTwo
WHEN f.state = com.package.State.THREE THEN f.deadlineThree
ELSE f.deadlineOne END")
Page<Foo> findSortedByNextDeadline(Pageable pageable);

Related

Finding a list of related objects by ID

Let's say for example I have a bridge table called PersonAnimal. I want to search for all the people who have a given animal's ID. The query so far looks like:
Animal animal = getById(Animal.class, animalId)
ObjectSelect
.query(PersonAnimal.class)
.where(PersonAnimal.ANIMAL.eq(animal))
.select(context)
However the first line in the above code segment shows that I first have to retrieve the related object from the database. I want to get rid of that database lookup and instead do something like:
ObjectSelect
.query(PersonAnimal.class)
.where(PersonAnimal.ANIMAL_ID.eq(animalId)) // <- Find by ID instead
.select(context)
Is that possible?
I am running version 4.1 of the Apache Cayenne ORM.
Just as I posted the question I found the answer. You need to create an Expression using a Property object like so:
val findByIdExpr = Property.create(PersonAnimal.ANIMAL.name, Long::class.java).eq(yourId)
val gotList = ObjectSelect
.query(PersonAnimal.class)
.where(findByIdExpr)
.select(context)
Above code is in Kotlin but is also easy to understand from a Java perspective.

Convert Integer to String in Spring Data JPA Query

I need to optimize a query that iterates over several objects and I wanted Spring Data to let the database handle it. I want to end up with a HashMap<String,String> that looks like
2134_9877, 9877
2134_2344, 2344
3298_9437, 9437
The SQL would be select convert(varchar,b.id)+'_'+convert(varchar,a.id)',a.id from t1 a join t2 b on a.jc = b.jc
So far, I've got Whatever-QL in the repository that looks like:
#Query("SELECT new map (a.bkey, a.akey) FROM mergeTable a WHERE a.discr= ?1")
The problem is, bkey is not unique, it is only unique when paired with akey and the monstrosity that I have to feed it to wants them combined with an underscore: 2345_2177.
I have tried a.bkey.toString and ''+a.bkey and new String(a.bkey) and just string(a.bkey) (that last gives a new exception but still doesn't work) but Spring doesn't like any of these. I can find no questions asking this and it seems I cannot use SQLServer's convert() function as this ain't SQL.
How can I concatenate the Integers as Strings with an underscore in this #Query?
PS: Using the native query that's been debugged in SQLServer throws some weird alias exception in Hibernate so I think 'going native' is predetermined to be a dead end.
If I have understood it right, the 'Whatever-QL' is called JPQL, and the operator CONCAT can be used. Only the use of it, as it accepts two or more parameters depends on the JPA version you are running.
Here is the answer.
JPA concat operator
You could add a getter to your entity like this:
public String getCombinedKey(){
return a.akey + "_" + a.bkey;
}
The advantage is you could handle here null's and other things if you want and it's more reusable in case you need this in another place. If you do it just in the repository you will have to copy it everytime.
Your query would then be:
#Query("SELECT new map (a.combinedKey, a.akey) FROM mergeTable a WHERE a.discr= ?1")

Hybris CategoryProductRelation type as ItemModel?

Quick question... so in Hybris, I have a query similar to this:
"SELECT {CPR:pk} FROM {CategoryProductRelation as CPR}, ...."
Basically, I need to extract the Product Code and Category Code from Java which I think are available as source / target respectively but my question is, just like there's ProductModel, CategoryModel, etc. is there anything like that for CategoryProductRelation?, probably something like a generic RelationModel to simply extract source / target and go from there?
You'll need to JOIN in the entities like this
SELECT {CPR:pk}, {c.code} FROM {CategoryProductRelation as CPR
JOIN Category AS c on {CPR.source} = {c.PK} } WHERE ...
Also, you can do that in the Service Layer by simply calling your query and accessing the properties right from the relation type:
..
CategoryProductRelationModel model = result.get(0)
String categoryCode = ((CategoryModel)model.getSource()).getCode()
Depending on your amount of data, this could be pretty ineffecient.

How to store all user activites in a website..?

I have a web application build in Django + Python that interact with web services (written in JAVA).
Now all the database management part is done by web-services i.e. all CRUD operations to actual database is done by web-services.
Now i have to track all User Activities done on my website in some log table.
Like If User posted a new article, then a new row is created into Articles table by web-services and side by side, i need to add a new row into log table , something like "User : Raman has posted a new article (with ID, title etc)"
I have to do this for all Objects in my database like "Article", "Media", "Comments" etc
Note : I am using PostgreSQL
So what is the best way to achieve this..?? (Should I do it in PostgreSQL OR JAVA ..??..And How..??)
So, you have UI <-> Web Services <-> DB
Since the web services talk to the DB, and the web services contain the business logic (i.e. I guess you validate stuff there, create your queries and execute them), then the best place to 'log' activities is in the services themselves.
IMO, logging PostgreSQL transactions is a different thing. It's not the same as logging 'user activities' anymore.
EDIT: This still means you create DB schema for 'logs' and write them to DB.
Second EDIT: Catching log worthy events in the UI and then logging them from there might not be the best idea either. You will have to rewrite logging if you ever decide to replace the UI, or for example, write an alternate UI for, say mobile devices, or something else.
For an audit table within the DB itself, have a look at the PL/pgSQL Trigger Audit Example
This logs every INSERT, UPDATE, DELETE into another table.
In your log table you can have various columns, including:
user_id (the user that did the action)
activity_type (the type of activity, such as view or commented_on)
object_id (the actual object that it concerns, such as the Article or Media)
object_type (the type of object; this can be used later, in combination with object_id to lookup the object in the database)
This way, you can keep track of all actions the users do. You'd need to update this table whenever something happens that you wish to track.
Whenever we had to do this, we overrode signals for every model and possible action.
https://docs.djangoproject.com/en/dev/topics/signals/
You can have the signal do whatever you want, from injecting some HTML into the page, to making an entry in the database. They're an excellent tool to learn to use.
I used django-audit-log and I am very satisfied.
Django-audit-log can track multiple models each in it's own additional table. All of these tables are pretty unified, so it should be fairly straightforward to create a SQL view that shows data for all models.
Here is what I've done to track a single model ("Pauza"):
class Pauza(models.Model):
started = models.TimeField(null=True, blank=False)
ended = models.TimeField(null=True, blank=True)
#... more fields ...
audit_log = AuditLog()
If you want changes to show in Django Admin, you can create an unmanaged model (but this is by no means required):
class PauzaAction(models.Model):
started = models.TimeField(null=True, blank=True)
ended = models.TimeField(null=True, blank=True)
#... more fields ...
# fields added by Audit Trail:
action_id = models.PositiveIntegerField(primary_key=True, default=1, blank=True)
action_user = models.ForeignKey(User, null=True, blank=True)
action_date = models.DateTimeField(null=True, blank=True)
action_type = models.CharField(max_length=31, choices=(('I', 'create'), ('U', 'update'), ('D', 'delete'),), null=True, blank=True)
pauza = models.ForeignKey(Pauza, db_column='id', on_delete=models.DO_NOTHING, default=0, null=True, blank=True)
class Meta:
db_table = 'testapp_pauzaauditlogentry'
managed = False
app_label = 'testapp'
Table testapp_pauzaauditlogentry is automatically created by django-audit-log, this merely creates a model for displaying data from it.
It may be a good idea to throw in some rude tamper protection:
class PauzaAction(models.Model):
# ... all like above, plus:
def save(self, *args, **kwargs):
raise Exception('Permission Denied')
def delete(self, *args, **kwargs):
raise Exception('Permission Denied')
As I said, I imagine you could create a SQL view with the four action_ fields and an additional 'action_model' field that could contain varchar references to model itself (maybe just the original table name).

Getting all users with a Role in Liferay

I'm new to Liferay development in general, so feel free to point out if I'm going about stuff totally the wrong way.
I'm trying to get a DynamicQuery object of all users within a certain group (I'll use this object to further filter another query I'll do against the message board). The User interface seems to have a roleIds property that I might be able to use, since I already know the roleId I'm interested in. But I can't find the proper way to query if roleIds contains a certain value.
Any ideas on what I want to do?
PS: I would have the exact SQL query I could ask directly, but I'd rather use Liferay's own connection pool, without needing to do some weird ext project thingy.
You don't need a DynamicQuery. These are the methods you are looking for in the classes that Dirk points out:
long[] UserServiceUtil.getRoleUserIds(long roleId)
or
long[] UserLocalServiceUtil.getRoleUserIds(long roleId)
List<User> UserLocalServiceUtil.getRoleUsers(long roleId)
Remember that the methods in the classes XXXLocalServiceUtil are not checking the permissions of the current user.
EDIT: If you are looking for all users with a given role within a given community:
long companyId= _X_; //Perhaps CompanyThreadLocal.getCompanyId() if you don't have it anywhere else?
Role role=RoleLocalServiceUtil.getRole(companyId, "Example Role");
Group group=GroupLocalServiceUtil.getGroup(companyId, "Example Community");
List<UserGroupRole> userGroupRoles = UserGroupRoleLocalServiceUtil.
getUserGroupRolesByGroupAndRole(groupId, role.getRoleId());
for(UserGroupRole userGroupRole:userGroupRoles){
User oneUser=userGroupRole.getUser();
}
The easiest way to access liferays own objects is by using the XXXServiceUtil classes (e.g. RoleServiceUtil.getUserRoles(userId)). Thus you rarely have to deal with any SQL directly. Either the RoleServiceUtil or UserServiceUtil might have what you need.
The roles of an Organizations are stored in the table UserGroupRole, so if you want to get the owner of an Organization you must use the following code:
boolean isOrgOwner =
UserGroupRoleLocalServiceUtil.hasUserGroupRole(
usr.getUserId(),
this.currentOrganization.getGroupId(),
RoleConstants.ORGANIZATION_OWNER);
If you want to retrieve all the Organization Owners of an organization:
List<User> administrators = new LinkedList<>();
List<UserGroupRole> allOrganizationAdministrators =
UserGroupRoleLocalServiceUtil.getUserGroupRolesByGroupAndRole(
this.currentOrganization.getGroupId(), roleId);
for (UserGroupRole userGroupRoleTemp : allOrganizationAdministrators) {
administrators.add(userGroupRoleTemp.getUser());
}
Cheers!

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