Jump Ref method D365FO

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Posted on by sylvesterPowerBi

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Dears,

i am overriding the JumpRef() method on my form using the below code

  public void jumpRef()
        {
            FormRun formRun;
            Args args = new Args();

            super();
            args.lookupTable(tableNum(XXX)); 
           

            args.lookupField(fieldNum(XXX,ZZ));
            

            args.lookupValue(XXX.ZZ);
            

            args.lookupRecord(XXX::find(XXX.ZZ));
            

            args.name(formStr(YYY)); 
            
            
            formRun = ClassFactory.formRunClass(args);
            formRun.init();
            formRun.run();
            formRun.wait();
            formRun.detach();
        }

the issue is when i click on the ID in the List form to open the new form, it doesn't select automatically the selected ID in the new form but always the new form is selecting the first ID.

i found out that the query range used on the new form is affecting the jumpref() cause after commenting the query range code everything worked just fine.

Any idea how to use Jumpref with QueryBuildRange ?

if any other requirement are necessary please tell me.

Regards,

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GirishS 27,825 Moderator on at

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RE: Jump Ref method D365FO

First you find out whether you are getting the caller menu item. Just add the info log in the execute query method to display caller menu item.

Also get the caller record in form init method. Declare table buffer globally and on the init method you can assign values like below.

TableName tableName1; //declare globally on form declaration.

Now override the init method and get the caller record.

tableName1 = element.args().record();

so that you can use that buffer in the executeQuery method to add range and values.

Thanks,

Girish S.

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waytod365 351 on at

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RE: Jump Ref method D365FO

Thanks for your replay Grish,

Below my code.

Args args = new Args(); args.record(TableA); MenuFunction menuFunction = new MenuFunction(menuitemDisplayStr(FormB), MenuItemType::Display); menuFunction.formViewOption(FormViewOption::Details); menuFunction.openMode(OpenMode::Edit); menuFunction.run(args); //Form B public void executeQuery() { if (element.args().menuItemName() == menuItemDisplayStr(FormA)) { this.queryBuildDataSource().addRange(fieldNum(TableB, RecId)).value(queryValue(element.args().record().RecId)); } super(); }

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GirishS 27,825 Moderator on at

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RE: Jump Ref method D365FO

If there is no direct relation between 2 tables you need to override the jump ref method.

Have you done that? Can you share a code with us.

Thanks,

Girish S.

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waytod365 351 on at

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RE: Jump Ref method D365FO

Hi, actually i have same issue but, there is no direct relation.

TableA and TableB

TableA has it refRecID and RefTableID.

Issue: it's always open first record from TableB. (FormB)

How to overcome this issue?

I did it code from FormB excuteQuery

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nmaenpaa 101,160 Moderator on at

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RE: Jump Ref method D365FO

Try this:

public void jumpRef() { Args args = new Args(); MenuFunction menuFunction = new MenuFunction(MenuItemDisplayStr(TheMenuItemOfTheFormYouWantToOpen), MenuItemType::Display); args.record(MyDataSourceName); args.caller(element); menuFunction.run(args); }

You should always use MenuFunction to launch forms, this way it takes into account security and only opens the form if your user has proper rights to open the form.

Also, make sure that you have defined table relations from your table to the table that is opened by the JumpRef. If the relations are set up correctly, the form should be filtered based on the caller record.

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Gunjan Bhattachayya 35,421 on at

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RE: Jump Ref method D365FO

Hi Sylvester,

In the jumpref Method, try declaring "args.record()"

In the form (YYY in your example) init() method, you can handle this condition with your QueryBuildRange.

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