如何在onDelete中获取当前的CoreData项目

时间:2020-06-02 13:46:23

标签: ios core-data swiftui

有两个实体ParentChild,这是一对多关系。一个父母和多个孩子。

我使用EditMode删除Child数据,例如:

@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)
                }
            }
        }
    }
}

deleting方法在另一个类中定义,例如:

public func deleting(at indexset: IndexSet) {

    CoreData.executeBlockAndCommit {

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

并且我还想在Parent发生时更新ChildonDelete实体的其他属性。 但是我必须找到已删除的当前Child项目。怎么做?

感谢您的帮助。

1 个答案:

答案 0 :(得分:1)

这里是可行的方法...(假设您的.loadDB返回数组,但通常类似的方法适用于任何随机访问集合)

通过Xcode 11.4(使用常规的项目数组)进行了测试

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)
    }
}