Exam MB-820 topic 1 question 36 discussion

In my opinion, if you look at the requirement result, it is sorted by Vendor No. first, and then Item No. if you want to arrange the result as requirement without using any property, the answer is dataitem(Vendor; Vendor) { column("Vendor No."; Vendor."Vendor No.") {} column("Vendor Name"; Vendor."Vendor Name") {} dataItem(Item;Item) { DataItemLink = "Vendor No." = Vendor."Vendor No."; SQLJoinType = InnerJoin; column("Item No."; Item."Item No.") {} column("Item Description"; Item."Item Description") {} } } noted ** if using data Item "Item" before "Vendor", you need to clarify orderBy property = Vendor No. and Item No.

dataitem(Vendor; Vendor){} dataitem(Item; Item){} SqlJoinType = InnerJoin;

Correct answer is 1 dataitem(Vendor; Vendor) 2 dataItem(Item;Item) 3 DataItemLink = "Vendor No." = Vendor."Vendor No.";

dataItem(Item;Item) { column("Item No."; Item."Item No.") {} column("Item Des"; Item."Item Des") {} column("Vendor No."; Item."Vendor No.") {} dataItem(Vendor;Vendor) { } }

The below example is a query that does what is requested, any code block of this exercise belong in this example. query 50000 "Items and Vendors" { DataItem(Item; "Item") { Column(ItemNumber; "No.") { } Column(Description; Description) { } Column(VendorNo; "Vendor No.") { } // Left Outer Join with the Vendor table, you want to show all item values, but show vendor name where vendor no. is available. DataItem(Vendor; "Vendor") { DataItemLink = "No." = Item."Vendor No."; SQLJoinType = LeftOuterJoin; Column(VendorName; Name) { } } } }