Invalid Relation between entities

Like (0) Share

Report

Posted on 5 Feb 2023 16:09:58 by D365FO user

465

I'm getting this error when exporting the composite entity: "Invalid relation between entities"

The relation between Entity2 and Entity1 is:
Entity1.Id = Entity2.Id

Entity1.JournalParmId= Entity2.JournalParmId

Entity1.LinkedJournalId = Entity2.JournalId

Entity1 EntityKey: JournalId

Entity2 EntityKey: JournalId

LinkedJouranId and LinkedJournalType are computed columns:

public static server str LinkedJournalId()
{

        return  strFmt('ISNULL((select top 1 JournalId from Table1 as table1 WHERE table1.PARTITION = %1 AND table1.DATAAREAID = %2 AND table1.JournalParmId = %3 AND table1.JournalId != %4 AND table1.Id = %5 AND table1.JournalType != %6), %7)'
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, Partition))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, DataAreaId))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, JournalParmId))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, JournalId))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, Id))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, JournalType))  // this is new
            , SysComputedColumn::comparisonLiteral('')
             );
    
}


public static server str LinkedJournalType()
    {

        return  strFmt('select top 1 JournalType from Table1 as table1 WHERE table1.PARTITION = %1 AND table1.DATAAREAID = %2 AND table1.JournalParmId = %3 AND table1.JournalId != %4 AND table1.RentId = %5 AND table1.JournalType != %6'
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, Partition))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, DataAreaId))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, JournalParmId))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, JournalId))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, Id))
            , SysComputedColumn::comparisonField(tableStr(Table1View), identifierStr(Table1), fieldStr(Table1, JournalType))
             );
}

*I've noticed that if JournalParmId in Entity 1 is not repeated then the export doesn't fail like example below

-- however, if more than JournalParmId in entity1, then it fails

All responses (13)

Answers (0)

D365FO user 465 on 16 Feb 2023 at 13:15:25

Like (0)

Report

RE: Invalid Relation between entities

Hi Martin,

I tried to do the entity in a different way without using computed column in relation: -- still same issue

I created a new link

|community.dynamics.com/.../error-when-exporting-the-entity-invalid-relation-between-entities
D365FO user 465 on 06 Feb 2023 at 12:02:50

Like (0)

Report

RE: Invalid Relation between entities

JournalParmId is when you create two Journals in the same button click.

So for example if you want to replace an item with another item, you click one button to do the replacements, where two journals gets created with the same JournalParmId and from a business point of view these two are linked --- this is an example of J2 and J3

And in the new case for (J4, J5 and J6) -- you can click a button to collect three items but they have a different address, so three journals gets created for each item with the same JournalParmId -- but those are not considered linked that's why i don't want them to appear in the linked entity (child)

However, if you collect three items that have the same address, only one journal will get created that contains these three items
Martin Dráb 231,723 Most Valuable Professional on 06 Feb 2023 at 11:49:47

Like (0)

Report

RE: Invalid Relation between entities

Sorry, but your data models no sense to me. For example, I got no information about the meaning of JournalParmId. I can't help you.

All I can say is that the design of your entities looks extremely weird, but I don't have enough to either understand it or to propose something more meaningful.
D365FO user 465 on 06 Feb 2023 at 11:12:29

Like (0)

Report
RE: Invalid Relation between entities

but those are different entities that are joined together in a composite entity. So if Entity1 (parent) which consists of table1 had J1, then Entity2 which consists of table1 as well can have J1 as well, as the entity key is JournalId for each one.

So table1 can have J1 and J1 can't be repeated true. But Entity1(Parent) can have J1 and Entity2(Child) can have J1

But anyways what I'm trying to do is Make Entity1 contain Journals that shouldn't appear in Entity2. That's why i made the link between them based on Entity1.LinkedJournalId = Entity2.JournalId

So i'm assuming if I join the two entities here based on ID JounralParmId and LinkedJournalId

The result should be like this -- when joining the two entities based on the relation i specified -- but i'm not sure why the entity is failing

Id JournalId JournalParmId JournalType LinkedJournalId LinkedJournalType Id JournalId JournalParmId JournalType 1 J1 1A 10 NULL NULL NULL NULL NULL 1 J3 2B 0 J2 15 1 J2 2B 15 1 J4 4C 15 NULL NULL NULL NULL NULL 1 J5 4C 15 NULL NULL NULL NULL NULL 1 J6 4C 15 NULL NULL NULL NULL NULL
Martin Dráb 231,723 Most Valuable Professional on 06 Feb 2023 at 10:36:24

Like (0)

Report

RE: Invalid Relation between entities

Yes, I know it's the table table - I'm talking about records in the table. When you say that JournalId in Table1 is unique and you have two records - a parent and a child - they must have two different Journal ID. If they had the same, JournalId wouldn't be unique.
D365FO user 465 on 06 Feb 2023 at 09:32:23

Like (0)

Report

RE: Invalid Relation between entities

Hey Martin,

But Parent and Child are exactly the same table which is table1

So table1 in general has J1,J2,J3,J4,J5,J6

-- J2 and J3 are linked and share the same JournalParmId

-- J4,J5,J6 are not linked from a business point of view but they still share the same journalParmId

So Parent Entity (which is table1 joined with another table) has J1,J3,J4,J5,J6 -- no J2

And I designed child entity in a way to show linked Journals from table1

so that the xml shows

J1 in Parent linked with nothing from child

J3 in Parent linked with j2 from child

J4 in Parent linked with nothing from child

J5 in Parent linked with nothing from child

J6 in Parent linked with nothing from child

So when i say child entity shouldn't show journals from ParentEntity.. is because child node should only show J2 in this case

that's why i made this link childEntity.JournalId = ParentEntity.LinkedJournalId

and they way i detected linkedJournalId in ParentEntity, is by looking at a journal who has the same JournalParmId but different JournalId -- So again J2 and J3 has same JournalParmId but because Parent has J2 i only want to get J3
Martin Dráb 231,723 Most Valuable Professional on 06 Feb 2023 at 08:54:38

Like (0)

Report

RE: Invalid Relation between entities

If Journal ID in the table is unique, two record (a parent and a child) clearly can't have the same Journal ID, because that wouldn't be unique.
D365FO user 465 on 06 Feb 2023 at 06:59:07

Like (0)

Report
RE: Invalid Relation between entities

Hi Martin,

Yes i need to check if JournalId is not equal to the parent. because Table1Entity and LinkedTable1Entity have the same table, so the inner node related to LinkedTable1Entity, should not show the same JournalId in the parent. And in the new case it should not show the same JournalId in the parent AND it should not show same JournalType as the parent

So currently here's the case that is breaking the entity as I mentioned before

Table1Entity // note that here there is no J2, because always when 2 records are actually linked, only 1 appears that i detect by joining table1 with another table
Id JournalId JournalParmId JournalType LinkedJournalId LinkedJournalType 1 J1 1A 10 NULL 1 J3 2B 0 J2 15 1 J4 4C 15 NULL 1 J5 4C 15 NULL 1 J6 4C 15 NULL

LinkedTable1Entity // contains table1 not joined with anything so J2 appears here, but because it's not in Parent entity I make sure it will only appear in inner node and not parent node.
Id JournalId JournalParmId JournalType 1 J1 1A 10 1 J2 2B 15 1 J3 2B 0 1 J4 4C 15 1 J5 4C 15 1 J6 4C 15

now when joining these two entities -- relation based on ID, JournalParmId and LinkedTable1Entity.JournalId = Table1Entity.LinkedJournalId

When I'm standing at J1, Linked node should show nothing

When I'm standing at J3, Linked node should show J2

When I'm standing at J4, Linked node should show nothing

When I'm standing at J5, Linked node should show nothing

When I'm standing at J6, Linked node should show nothing

** my previous comment that contains the xml shows all the details as well

goshoom when I had J1,J2 and J3 in table1, the entity didn't break, because when J2 and J3 where linked, EntityTable1 didn't contain both J2 and J3, just J3.

But now since there is J4.J5,J6 and they both appear in both entities by sharing the same journalParmId, but they are not actually linked,it fails.
So somehow it's failing when the JournalParmId is repeated in Parent Entity. but why??
Martin Dráb 231,723 Most Valuable Professional on 06 Feb 2023 at 06:27:11

Like (0)

Report

RE: Invalid Relation between entities

If Journal ID is unique, involving the other fields is useless, because they can't make it any more unique.

Also, you don't need to do anything to make sure that "its JournalId is not equal to JournalId of the parent". If Journal ID is unique, a record can't have the Journal ID same as any other journal, including the parent.
D365FO user 465 on 05 Feb 2023 at 19:06:16

Like (0)

Report

RE: Invalid Relation between entities

Hi Martin,

No JournalId alone is unique -- and the primary key of the table is JournalId

ID is there just to be more sure -- ID is like the order number

so you can say as well: inner node needs to display all records that have the same ID and same JournalParmId as the parent but it's JournalId is not equal to JournalId of the parent.

goshoom but after the new case, inner node needs to display all records that have the same ID and same JournalParmId as the parent but it's JournalId is not equal to JournalId of the parent and it's JournalType is not equal to JournalType of the parent

Community site session details

Invalid Relation between entities

Helpful resources

Quick Links

Daivat Vartak – Community Spotlight

Announcing Our 2025 Season 1 Super Users!

Kudos to the February Top 10 Community Stars!

Leaderboard

Featured topics

Product updates

Community site session details

Invalid Relation between entities

Helpful resources

Quick Links

Daivat Vartak – Community Spotlight

Announcing Our 2025 Season 1 Super Users!

Kudos to the February Top 10 Community Stars!

Subscribe to this forum!

Select categories

Leaderboard

Featured topics

Product updates