更新数组索引但它更改了 IOS 中的所有索引
update array index but its changing all index in IOS
我在这里尝试更新索引中的特定值,但它更新了所有索引
class salarty {
var sal_id :String = ""
var sal_amount :String = ""
var sal_iScreated :Bool = false
init(sal_id : String, sal_amount : String, sal_iScreated: Bool){
self.sal_id = sal_id
self.sal_amount = sal_id
self.sal_iScreated = sal_id
}
}
class Employee {
let id: Int, firstName: String, lastName: String
var salaryArray :[salarty] = [salarty]()
init(id: Int, firstName: String, lastName: String, salaryArray: [salarty]) {
self.id = id
self.firstName = firstName
self.lastName = lastName
self.salaryArray = salaryArray
}
}
array_SalaryDetails.append(salarty(sal_id : "2", sal_amount : "3000", sal_iScreated: false))
let employeeArray = [
Employee(id: 1, firstName: "Jon", lastName: "Skeet",salaryArray :array_SalaryDetails),
Employee(id: 2, firstName: "Darin", lastName: "Dimitrov",salaryArray :array_SalaryDetails),
Employee(id: 4, firstName: "Hans", lastName: "Passant",salaryArray :array_SalaryDetails)
]
我在这里尝试了一些代码,但它不起作用
方法一:
var model = Employee[(indexPath?.section)!].salaryArray
let model2 = model[(indexPath?.row)!]
model = model.map{
let mutableval = [=12=]
if [=12=].sal_id == model2.sal_id {
mutableBook.sal_iScreated = !model2.sal_iScreated
}
return mutableval
}
方法二:
let model = array_Main[(indexPath?.section)!].serviceArray
let model2 = model[(indexPath?.row)!]
model2.service_isSelected = !model2.service_isSelected
那是因为在 class 的 Swift
实例中 通过引用 而不是 通过值 传递。
因此,您创建并传递给多个 Employee 对象的工资数组最终引用了相同的工资对象。因此,更改任何 Employee 对象中任何数组中的薪水值都会反映所有 Employee 对象的变化
两种可能的解决方案:
方案一:首选方案
将 salarty
更改为 struct
而不是 class
并且如果可能的话也更改 salary 的拼写 :P
struct salarty {
var sal_id :String = ""
var sal_amount :String = ""
var sal_iScreated :Bool = false
init(sal_id : String, sal_amount : String, sal_iScreated: Bool){
self.sal_id = sal_id
self.sal_amount = sal_id
self.sal_iScreated = sal_iScreated
}
}
方案二:
在分配之前为每个员工创建薪水数组的深层副本。
class salarty : NSObject, NSCopying {
func copy(with zone: NSZone? = nil) -> Any {
let copy = salarty(sal_id: self.sal_id, sal_amount: self.sal_amount, sal_iScreated: self.sal_iScreated)
return copy
}
var sal_id :String = ""
var sal_amount :String = ""
var sal_iScreated :Bool = false
init(sal_id : String, sal_amount : String, sal_iScreated: Bool){
self.sal_id = sal_id
self.sal_amount = sal_id
self.sal_iScreated = sal_iScreated
}
}
对于每个员工,您可以使用以下任一方法创建一个新数组
array_SalaryDetails.append(salarty(sal_id : "2", sal_amount : "3000", sal_iScreated: false))
//salary array copy for first employee
var firstEmployeearrayCopy = [salarty]()
for salary in array_SalaryDetails {
firstEmployeearrayCopy.append(salary.copy() as! salarty)
}
Employee(id: 1, firstName: "Jon", lastName: "Skeet",salaryArray : firstEmployeearrayCopy)
//salary array copy for second employee
var secondEmployeearrayCopy = [salarty]()
for salary in array_SalaryDetails {
secondEmployeearrayCopy.append(salary.copy() as! salarty)
}
Employee(id: 2, firstName: "Darin", lastName: "Dimitrov",salaryArray :secondEmployeearrayCopy)
或者您可以为每个员工再次创建数组 n。
In swift Array 是 通过值 传递的,但是数组中的对象(class 的实例)仍将通过引用传递,因此您将不得不执行所有这些解决方法:)
一条忠告:
Class 名称应以大写字母开头。只有变量名遵循驼峰式大小写。
希望对您有所帮助
我在这里尝试更新索引中的特定值,但它更新了所有索引
class salarty {
var sal_id :String = ""
var sal_amount :String = ""
var sal_iScreated :Bool = false
init(sal_id : String, sal_amount : String, sal_iScreated: Bool){
self.sal_id = sal_id
self.sal_amount = sal_id
self.sal_iScreated = sal_id
}
}
class Employee {
let id: Int, firstName: String, lastName: String
var salaryArray :[salarty] = [salarty]()
init(id: Int, firstName: String, lastName: String, salaryArray: [salarty]) {
self.id = id
self.firstName = firstName
self.lastName = lastName
self.salaryArray = salaryArray
}
}
array_SalaryDetails.append(salarty(sal_id : "2", sal_amount : "3000", sal_iScreated: false))
let employeeArray = [
Employee(id: 1, firstName: "Jon", lastName: "Skeet",salaryArray :array_SalaryDetails),
Employee(id: 2, firstName: "Darin", lastName: "Dimitrov",salaryArray :array_SalaryDetails),
Employee(id: 4, firstName: "Hans", lastName: "Passant",salaryArray :array_SalaryDetails)
]
我在这里尝试了一些代码,但它不起作用
方法一:
var model = Employee[(indexPath?.section)!].salaryArray
let model2 = model[(indexPath?.row)!]
model = model.map{
let mutableval = [=12=]
if [=12=].sal_id == model2.sal_id {
mutableBook.sal_iScreated = !model2.sal_iScreated
}
return mutableval
}
方法二:
let model = array_Main[(indexPath?.section)!].serviceArray
let model2 = model[(indexPath?.row)!]
model2.service_isSelected = !model2.service_isSelected
那是因为在 class 的 Swift
实例中 通过引用 而不是 通过值 传递。
因此,您创建并传递给多个 Employee 对象的工资数组最终引用了相同的工资对象。因此,更改任何 Employee 对象中任何数组中的薪水值都会反映所有 Employee 对象的变化
两种可能的解决方案:
方案一:首选方案
将 salarty
更改为 struct
而不是 class
并且如果可能的话也更改 salary 的拼写 :P
struct salarty {
var sal_id :String = ""
var sal_amount :String = ""
var sal_iScreated :Bool = false
init(sal_id : String, sal_amount : String, sal_iScreated: Bool){
self.sal_id = sal_id
self.sal_amount = sal_id
self.sal_iScreated = sal_iScreated
}
}
方案二:
在分配之前为每个员工创建薪水数组的深层副本。
class salarty : NSObject, NSCopying {
func copy(with zone: NSZone? = nil) -> Any {
let copy = salarty(sal_id: self.sal_id, sal_amount: self.sal_amount, sal_iScreated: self.sal_iScreated)
return copy
}
var sal_id :String = ""
var sal_amount :String = ""
var sal_iScreated :Bool = false
init(sal_id : String, sal_amount : String, sal_iScreated: Bool){
self.sal_id = sal_id
self.sal_amount = sal_id
self.sal_iScreated = sal_iScreated
}
}
对于每个员工,您可以使用以下任一方法创建一个新数组
array_SalaryDetails.append(salarty(sal_id : "2", sal_amount : "3000", sal_iScreated: false))
//salary array copy for first employee
var firstEmployeearrayCopy = [salarty]()
for salary in array_SalaryDetails {
firstEmployeearrayCopy.append(salary.copy() as! salarty)
}
Employee(id: 1, firstName: "Jon", lastName: "Skeet",salaryArray : firstEmployeearrayCopy)
//salary array copy for second employee
var secondEmployeearrayCopy = [salarty]()
for salary in array_SalaryDetails {
secondEmployeearrayCopy.append(salary.copy() as! salarty)
}
Employee(id: 2, firstName: "Darin", lastName: "Dimitrov",salaryArray :secondEmployeearrayCopy)
或者您可以为每个员工再次创建数组 n。
In swift Array 是 通过值 传递的,但是数组中的对象(class 的实例)仍将通过引用传递,因此您将不得不执行所有这些解决方法:)
一条忠告:
Class 名称应以大写字母开头。只有变量名遵循驼峰式大小写。
希望对您有所帮助