如何在CoreData中获取当前onDelete项
如何在CoreData中获取当前onDelete项
提问于 2020-06-02 13:46:23
有两个实体Parent和Child，它们是一对多的关系。一个父母多个孩子。
我使用EditMode删除Child数据，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，如下所示：
public func deleting(at indexset: IndexSet) {
CoreData.executeBlockAndCommit {
for index in indexset {
CoreData.stack.context.delete(self.fetchedObjects[index])
}
}
}我还想在发生Parent和Child实体时更新onDelete的其他属性。但是，我必须找到当前的Child 项目，该项目被删除了。怎么做？
谢谢你的帮助。
回答 1
Stack Overflow用户
回答已采纳
发布于 2020-06-02 14:27:37
这是可能的方法..。(假设您的.loadDB返回数组，但通常类似于任何随机访问集合)
用Xcode 11.4测试(使用常规项数组)
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)
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/62153505
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