How do I use a Fetch Request for multiple entities in CoreData?

Question

Goal: I want to to show a list of Todo items by their due dates which I have already achieved. But I also wanna show a list of a different entity, Categories on the same View above the list of Todo items. The categories are buttons to take you to a list of todo items filtered to that category. I set a relationship of Category to have many todos. How do I change my FetchRequest to support the added relationship?

Here is the current SectionedFetchRequest below. If I try to add a new FetchRequest for Categories I get a crash.

    @SectionedFetchRequest(entity: Todo.entity(),
                           sectionIdentifier: \.dueDateRelative,
                           sortDescriptors: [NSSortDescriptor(keyPath: \Todo.dueDate, ascending: true)],
                           predicate: nil,
                           animation: Animation.linear)
    var sections: SectionedFetchResults<String, Todo>

                    ForEach(sections) { section in
                        Section(header: Text(section.id.description)) {
                            ForEach(section) { todo in
                                TodoRowView(todo: todo)
                                    .frame(maxWidth: .infinity)
                                    .listRowSeparator(.hidden)
                            }
                            .onDelete { indexSet in
                                deleteTodo(section: Array(section), offsets: indexSet)
                            }
                        }
                    }

// Causes Crash when added to existing Fetch Request
//@FetchRequest(entity: Category.entity(), sortDescriptors: []) var categories: FetchedResults<Category>

What do you want to do with it? Show it above or below? you can access the `Category` by using `todo.relationship`. Why do you need a second `FetchRequest`? Your question/goal is unclear. — lorem ipsum, Dec 24 '21 at 00:26
I want to add sections above the Todo List for different `Categories`. Tapping the category would open a new list with only `Todos` in that category. ("Inbox, Work, Personal") They can select a category to add the todo to from the form. @loremipsum — Justin Comstock, Dec 24 '21 at 00:42

score 0 · Accepted Answer · answered Dec 24 '21 at 02:25

You might be overthinking this. If you add a sectionIdentifier for the category you can offer the option relatively easy.

The variable would look like this

extension Todo{
    @objc
    var categoryTitle: String{
        self.relationship?.title ?? "no category"
    }
}

Then add a few variables to the View

//Dictionary to store sort options
let sortOptions: [String: KeyPath<Todo, String>] = ["due date": \Todo.dueDateRelative, "category":\Todo.categoryTitle]
//Variable to use to filter list
@State var selectedSection: String = ""
//filter the `sections` by the selected section
//You can use nsPredicate too
var filteredSections: [SectionedFetchResults<String, Todo>.Element] {
    sections.filter({ val in
        if !selectedSection.isEmpty {
            return val.id == selectedSection
        }else{
            return true
        }
    })
}

Then give the user some buttons to select

//Give the user sort options
Menu("sort"){
    ForEach(Array(sortOptions.keys).sorted(by: <), id:\.self, content: { key in
        Button(key, action: {
            sections.sectionIdentifier = sortOptions[key]!
            selectedSection = ""
        })
    })
}
//Give the user section options
Picker("sections", selection: $selectedSection, content: {
    ForEach(sections, content: {section in
        Text(section.id).tag(section.id)
    })
    Text("all").tag("")
}).pickerStyle(.segmented)

The ForEach would display the user selections automatically

ForEach(filteredSections) { section in

This helped but I want to use a TabView, I have made a new question with this issue. https://stackoverflow.com/questions/70476155/how-to-create-dynamic-tabview-filtered — Justin Comstock, Dec 24 '21 at 19:52

How do I use a Fetch Request for multiple entities in CoreData?

1 Answers1