在mysql中返回数组选择

Question

I'm trying to group inventory results by the model and manufacturer name, display the amount of items matching and 1 result per grouping. With that said, I'd like to try and retrieve all inventory id's within the group. Wondering if this is possible... Any ideas?

FYI - I'm using Laravel, the line in question is the ->selectRaw(CambridgeID as CambridgeIDArray)

$getMatchingInventory = DB::table('inventory')
                                ->selectRaw('*, count(*) as groupTotal')
                                ->whereRaw("MATCH(ManufacturerNameMatch, SubCategoryNameMatch, MainCategoryNameMatch, Model_Name, Title_Override, Description_Old) AGAINST ('$final' IN BOOLEAN MODE)")
                                ->selectRaw('CambridgeID as CambridgeIDArray')
                                ->groupBy('Model_Name', 'ManufacturerNameMatch')
                                ->having('Units_OnHand', '>=', '1')
                                ->orderBy('ManufacturerNameMatch')
                                //->paginate(15);
                                ->get();

Answer 1

try this

$getMatchingInventory = DB::table('inventory')
                            ->select(DB::raw("GROUP_CONCAT(CambridgeID) as `CambridgeIDArray`, count(*) as `groupTotal`"))
                            ->whereRaw("MATCH(ManufacturerNameMatch, SubCategoryNameMatch, MainCategoryNameMatch, Model_Name, Title_Override, Description_Old) AGAINST ('$final' IN BOOLEAN MODE)")
                            ->groupBy('Model_Name', 'ManufacturerNameMatch')
                            ->having('Units_OnHand', '>=', '1')
                            ->orderBy('ManufacturerNameMatch') 
                            ->get();

Answer 2

You should be able to use GROUP_CONCAT for that, see: http://dev.mysql.com/doc/refman/5.7/en/group-by-functions.html#function_group-concat

You can use specify output format options (eg, SEPARATOR) and use additional string manipulation as needed within the GROUP_CONCAT .

(Fyi, using raw MySQL, at least for this question, would make it easier to parse.)