如何在onDelete中获取当前的CoreData项。

Question

有两个实体 Parent 和 Child这是一对多的关系。一个父母和多个孩子。

我使用的是 EditMode 删去 Child 等数据。

@ObservedObject private var db = CoreDataDB<Child>(predicate: "parent")

var body: some View {

    VStack {
        Form {

            Section(header: Text("Title")) {
                ForEach(db.loadDB(relatedTo: parent)) { child in

                    if self.editMode == .active {
                        ChildListCell(name: child.name, order: child.order)
                    } else {
                        ...
                    }
                }
                .onDelete { indices in 
                  // how to know which child is by indices here ???
                  let thisChild = child // <-- this is wrong!!!

                  self.db.deleting(at: indices)
                }
            }
        }
    }
}

该 deleting 方法被定义在另一个类中，比如。

public func deleting(at indexset: IndexSet) {

    CoreData.executeBlockAndCommit {

        for index in indexset {
            CoreData.stack.context.delete(self.fetchedObjects[index])
        }
    }
}

而我也想更新其他的Attribute of Parent 和 Child 实体时 onDelete 发生。但我必须找到当前的 Child 被删除的项目. 怎么做呢？

谢谢大家的帮助。

Answer 1

这里是一个可能的方法......（假设你的.loadDB返回数组，但一般来说，类似的方法将适用于任何随机访问集合

用Xcode 11.4测试（使用常规的项目数组）。

var body: some View {
    VStack {
        Form {
            Section(header: Text("Title")) {
                // separate to standalone function...
                self.sectionContent(with: db.loadDB(relatedTo: parent))
            }
        }
    }
}

// ... to have access to shown container
private func sectionContent(with children: [Child]) -> some View {
    // now we have access to children container in internal closures
    ForEach(children) { child in

        if self.editMode == .active {
            ChildListCell(name: child.name, order: child.order)
        } else {
            ...
        }
    }
    .onDelete { indices in

        // children, indices, and child by index are all valid here
        if let first = indices.first {
            let thisChild = children[first]       // << here !!
            // do something here
        }

        self.db.deleting(at: indices)
    }
}