Phi*_*nie 11 xcode ios parse-platform swift
我搜索过,但是我找不到熟悉的答案,所以......
我将编写一个类来处理解析方法,如更新,添加,提取和删除.
func updateParse(className:String, whereKey:String, equalTo:String, updateData:Dictionary<String, String>) {
let query = PFQuery(className: className)
query.whereKey(whereKey, equalTo: equalTo)
query.findObjectsInBackgroundWithBlock {(objects, error) -> Void in
if error == nil {
//this will always have one single object
for user in objects! {
//user.count would be always 1
for (key, value) in updateData {
user[key] = value //Cannot assign to immutable expression of type 'AnyObject?!'
}
user.saveInBackground()
}
} else {
print("Fehler beim Update der Klasse \(className) where \(whereKey) = \(equalTo)")
}
}
}
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由于我现在即将学习迅速,我很乐意通过一点宣言得到答案,这样我就可以学到更多.
顺便说一句:我后来称这个方法是这样的:
parseAdd.updateParse("UserProfile", whereKey: "username", equalTo: "Phil", updateData: ["vorname":self.vornameTextField!.text!,"nachname":self.nachnameTextField!.text!,"telefonnummer":self.telefonnummerTextField!.text!])
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Zpa*_*bor 34
在swift中,很多类型被定义为structs,默认情况下是不可变的.
我这样做有同样的错误:
protocol MyProtocol {
var anInt: Int {get set}
}
class A {
}
class B: A, MyProtocol {
var anInt: Int = 0
}
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在另一个班级:
class X {
var myA: A
...
(self.myA as! MyProtocol).anInt = 1 //compile error here
//because MyProtocol can be a struct
//so it is inferred immutable
//since the protocol declaration is
protocol MyProtocol {...
//and not
protocol MyProtocol: class {...
...
}
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所以一定要有
protocol MyProtocol: class {
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在做这样的铸造时
vad*_*ian 10
错误消息说,您正在尝试更改不可变对象,这是不可能的.
默认情况下,声明为方法参数或闭包中的返回值的对象是不可变的.
要使对象可变,要么var在方法声明中添加关键字,要么添加一行来创建可变对象.
默认情况下,重复循环中的索引变量也是不可变的.
在这种情况下,插入一行以创建可变副本,并将索引变量声明为可变.
在枚举时要小心更改对象,这可能会导致意外行为
...
query.findObjectsInBackgroundWithBlock {(objects, error) -> Void in
if error == nil {
//this will always have one single object
var mutableObjects = objects
for var user in mutableObjects! {
//user.count would be always 1
for (key, value) in updateData {
user[key] = value
...
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