I am getting the error message Cannot assign value of type 'String?' to type 'Int'
I have browsed through other questions like this but it still shows the error.
if sunscreenName.text != nil && reapplyTime.text != nil {
sunscreen = sunscreenName.text!
reApplyTime = reapplyTime.text
//Some sort of message such as Progress hud
}
Thanks in advance!
I got your problem, actually what happens here Swift is is type safe langauge
So what you are doing is is to store a String value in Int which will not happen automatically you need to convert it to Int
like this
Int(sunscreenName.text)
But there is a catch there not all string are convertible to Int type, fo e.g.
let name = "roshan"
if you try to convert it to Int it will give you a nil
let a = Int(name)
So its better you do a optional Binding here provided by Swift
if let sunValue = Int(sunscreenName.text),let reApplyValue = Int(reapplyTime.text) {
sunscreen = sunValue
reApplyTime = reApplyValue
}
I recommend reading through The Swift Programming Language to get a better understanding of Swift and its fundamental concepts, since this question is fairly basic.
You make several mistakes:
if sunscreenName.text != nil && reapplyTime.text != nil {
This is wrong. In Swift, if you plan to use the value later, you should use if let rather than comparing to nil. Comparing to nil leaves the values optional, but if let unwraps them. So, do this instead:
if let sunscreenText = sunscreenName.text, let reapplyText = reapplyTime.text {
Now you have the sunscreenText and reapplyText variables, which are typed String, not String? (i.e. they are not optional).
Now, there's these two lines.
sunscreen = sunscreenName.text!
reApplyTime = reapplyTime.text
You don't say which one is giving the error, but the issue is the same in either case. First, use our unwrapped sunscreenText and reapplyText variables instead of sunscreenName.text! and reapplyTime.text. Next, if one of these is meant to be an Int instead of a String, cast it. Swift is not like JavaScript, in that it won't automatically convert values from one type to another, so if something is a string and we need an integer, we have to convert it ourselves.
(assuming reapplyTime was the line that was giving the error:)
if let reapplyInt = Int(reapplyText) {
reapplyTime = reapplyInt
}
The reason we have to unwrap is because Int(String) can return nil if the string is something that can't be converted to an integer. Alternately, we could just provide a default value:
reapplyTime = Int(reapplyText) ?? 0 // sets to 0 if it can't parse the string as an integer
I have a case where when my viewControler starts in viewDidLoad I have to load some data using NSUserDefaults.standardUserDefaults() which doesn't exist in this monent. This data are saved when I tap send Button in the same viewController and I need this data when I open this viewController again. Now it looks like that:
var orderHistory = [String:String]()
vievDidLoad(){
let userDefault = NSUserDefaults.standardUserDefaults()
let orderHistory = userDefault.objectForKey("orderHistory")
if orderHistory == nil {
self.orderHistory = orderHistory["name":"", "surname":""] as! [String:String]
} else {
self.orderHistory = orderHistory as! [String:String]
{
}// end viewDidLoad
In this moment I recieve an imformation, I have a problem with memory. How should I avoid this situation?
As Leo Dabus said you should try using the ?? nil coalescing operator.
ObjectForKey does not provide a default value because it doesnt know what kind of object it is until you set it the first time. This results in a nil crash if you try to access it value without having it set once.
Compare this to say "boolForKey" where you dont have to do this, because it knows you are dealing with boolean values and therefore defaults to false automatically.
You also dont have to create 2 orderHistory dictionaries, it just makes your code more confusing.
Try this instead
var orderHistory = [String:String]()
vievDidLoad(){
let userDefault = NSUserDefaults.standardUserDefaults()
orderHistory = userDefault.objectForKey("orderHistory") as? [String: String] ?? orderHistory
//than just use the 1 dictionary without the if statements or creating another one.
}// end viewDidLoad
You check if saved data exists (as? [String: String]) and update the dictionary accordingly. If no saved data exists it will use the default values in orderHistory (?? orderHistory), which in your case is an empty dictionary.
This way you dont have to do a nil check, its all done in that one line.
Also try putting your keys into structs or global files so that you avoid typos. I see people not doing this all the time and its really bad practice.
So for example, above your class create a struct
struct Key {
static let orderHistory = "OrderHistory"
}
and use it like so
...objectForKey(Key.orderHistory)
This code makes no sense:
if orderHistory == nil
{
self.orderHistory = orderHistory["name":"", "surname":""] as! [String:String]
}
The if statement guarantees that orderHistory is nil, thereby guaranteeing that the attempt to fetch keys from orderHistory will crash. Actually, that doesn't look like valid Swift. I would expect that line to throw a compiler error.
Are you trying to create a new dictionary?
If so, your code should read like this:
if orderHistory == nil
{
self.orderHistory = ["name":"", "surname":""]
}
I am using a keychain wrapper written in Swift.
When value is nil, this wrapper I'm using automatically saves data as "nil" instead of saving as an actual nil value.
For ex, ["last_name" : "nil"].
In one of my label in my iOS app, I'm returning last_name value.
It bothers me that a text in label is saying "nil" when there supposed to be nothing.
I remember in Swift that there is a syntax like A ?? B which puts B instead if A is not valid. But I cannot remember it right now.
In my app, I want to something like below:
if the value I'm looking for is nil, then input "" instead of "nil"
If keychain["last_name"] == nil ?? ""
I know this is a horrible explanation but this is all I could come up with.
?? is the nil coalescing operator. It is used for unwrapping optionals — a ?? b is shorthand for a != nil ? a! : b.
You probably want to use the ternary operator instead. You could even combine them like this:
let lastName = keychain["last_name"] == "nil" ? "" : (keychain["last_name"] ?? "")
That said, it might be easier to just filter out these responses immediately after getting the keychain instead of checking every time:
var keychain = ["first_name" : "aaron", "last_name" : "nil"]
keychain.forEach { if $0.1 == "nil" { keychain.removeValueForKey($0.0) } }
print(keychain) //["first_name": "aaron"]
The $0 is a shorthand argument name. Here's the long form version of the same code:
keychain.forEach { (key, value) -> () in
if value == "nil" {
keychain.removeValueForKey(key)
}
}
I am doing what I believe to be a very simple task. I'm trying to get a value out of a dictionary if the key exists. I am doing this for a couple keys in the dictionary and then creating an object if they all exist (basically decoding a JSON object). I am new to the language but this seems to me like it should work, yet doesn't:
class func fromDict(d: [String : AnyObject]!) -> Todo? {
let title = d["title"]? as? String
// etc...
}
It gives me the error: Operand of postfix ? should have optional type; type is (String, AnyObject)
HOWEVER, if I do this, it works:
class func fromDict(d: [String : AnyObject]!) -> Todo? {
let maybeTitle = d["title"]?
let title = maybeTitle as? String
// etc...
}
It appears to be basic substitution but I may be missing some nuance of the language. Could anyone shed some light on this?
The recommended pattern is
if let maybeTitle = d["title"] as? String {
// do something with maybeTitle
}
else {
// abort object creation
}
It is possibly really a question of nuance. The form array[subscript]? is ambiguous because it could mean that the whole dictionary (<String:AnyObject>) is optional while you probably mean the result (String). In the above pattern, you leverage the fact that Dictionary is designed to assume that accessing some key results in an optional type.
After experimenting, and noticing that the ? after as is just as ambiguous, more, here is my solution:
var dictionary = ["one":"1", "two":"2"]
// or var dictionary = ["one":1, "two":2]
var message = ""
if let three = dictionary["three"] as Any? {
message = "\(three)"
}
else {
message = "No three available."
}
message // "No three available."
This would work with all non-object Swift objects, including Swift Strings, numbers etc. Thanks to Viktor for reminding me that String is not an object in Swift. +
If you know the type of the values you can substitute Any? with the appropriate optional type, like String?
There are a few of things going on here.
1) The ? in d["title"]? is not correct usage. If you're trying to unwrap d["title"] then use a ! but be careful because this will crash if title is not a valid key in your dictionary. (The ? is used for optional chaining like if you were trying to call a method on an optional variable or access a property. In that case, the access would just do nothing if the optional were nil). It doesn't appear that you're trying to unwrap d["title"] so leave off the ?. A dictionary access always returns an optional value because the key might not exist.
2) If you were to fix that:
let maybeTitle = d["title"] as? String
The error message changes to: error: '(String, AnyObject)' is not convertible to 'String'
The problem here is that a String is not an object. You need to cast to NSString.
let maybeTitle = d["title"] as? NSString
This will result in maybeTitle being an NSString?. If d["title"] doesn't exist or if the type is really NSNumber instead of NSString, then the optional will have a value of nil but the app won't crash.
3) Your statement:
let title = maybeTitle as? String
does not unwrap the optional variable as you would like. The correct form is:
if let title = maybeTitle as? String {
// title is unwrapped and now has type String
}
So putting that all together:
if let title = d["title"] as? NSString {
// If we get here we know "title" is a valid key in the dictionary, and
// we got the type right. title has now been unwrapped and is ready to use
}
title will have the type NSString which is what is stored in the dictionary since it holds objects. You can do most everything with NSString that you can do with String, but if you need title to be a String you can do this:
if var title:String = d["title"] as? NSString {
title += " by Poe"
}
and if your dictionary has NSNumbers as well:
if var age:Int = d["age"] as? NSNumber {
age += 1
}
I'm using Xcode 6 Beta 4. I have this weird situation where I cannot figure out how to appropriately test for optionals.
If I have an optional xyz, is the correct way to test:
if (xyz) // Do something
or
if (xyz != nil) // Do something
The documents say to do it the first way, but I've found that sometimes, the second way is required, and doesn't generate a compiler error, but other times, the second way generates a compiler error.
My specific example is using the GData XML parser bridged to swift:
let xml = GDataXMLDocument(
XMLString: responseBody,
options: 0,
error: &xmlError);
if (xmlError != nil)
Here, if I just did:
if xmlError
it would always return true. However, if I do:
if (xmlError != nil)
then it works (as how it works in Objective-C).
Is there something with the GData XML and the way it treats optionals that I am missing?
In Xcode Beta 5, they no longer let you do:
var xyz : NSString?
if xyz {
// Do something using `xyz`.
}
This produces an error:
does not conform to protocol 'BooleanType.Protocol'
You have to use one of these forms:
if xyz != nil {
// Do something using `xyz`.
}
if let xy = xyz {
// Do something using `xy`.
}
To add to the other answers, instead of assigning to a differently named variable inside of an if condition:
var a: Int? = 5
if let b = a {
// do something
}
you can reuse the same variable name like this:
var a: Int? = 5
if let a = a {
// do something
}
This might help you avoid running out of creative variable names...
This takes advantage of variable shadowing that is supported in Swift.
Swift 3.0, 4.0
There are mainly two ways of checking optional for nil. Here are examples with comparison between them
1. if let
if let is the most basic way to check optional for nil. Other conditions can be appended to this nil check, separated by comma. The variable must not be nil to move for the next condition. If only nil check is required, remove extra conditions in the following code.
Other than that, if x is not nil, the if closure will be executed and x_val will be available inside. Otherwise the else closure is triggered.
if let x_val = x, x_val > 5 {
//x_val available on this scope
} else {
}
2. guard let
guard let can do similar things. It's main purpose is to make it logically more reasonable. It's like saying Make sure the variable is not nil, otherwise stop the function. guard let can also do extra condition checking as if let.
The differences are that the unwrapped value will be available on same scope as guard let, as shown in the comment below. This also leads to the point that in else closure, the program has to exit the current scope, by return, break, etc.
guard let x_val = x, x_val > 5 else {
return
}
//x_val available on this scope
One of the most direct ways to use optionals is the following:
Assuming xyz is of optional type, like Int? for example.
if let possXYZ = xyz {
// do something with possXYZ (the unwrapped value of xyz)
} else {
// do something now that we know xyz is .None
}
This way you can both test if xyz contains a value and if so, immediately work with that value.
With regards to your compiler error, the type UInt8 is not optional (note no '?') and therefore cannot be converted to nil. Make sure the variable you're working with is an optional before you treat it like one.
From swift programming guide
If Statements and Forced Unwrapping
You can use an if statement to find out whether an optional contains a
value. If an optional does have a value, it evaluates to true; if it
has no value at all, it evaluates to false.
So the best way to do this is
// swift > 3
if xyz != nil {}
and if you are using the xyz in if statement.Than you can unwrap xyz in if statement in constant variable .So you do not need to unwrap every place in if statement where xyz is used.
if let yourConstant = xyz {
//use youtConstant you do not need to unwrap `xyz`
}
This convention is suggested by apple and it will be followed by devlopers.
Although you must still either explicitly compare an optional with nil or use optional binding to additionally extract its value (i.e. optionals are not implicitly converted into Boolean values), it's worth noting that Swift 2 has added the guard statement to help avoid the pyramid of doom when working with multiple optional values.
In other words, your options now include explicitly checking for nil:
if xyz != nil {
// Do something with xyz
}
Optional binding:
if let xyz = xyz {
// Do something with xyz
// (Note that we can reuse the same variable name)
}
And guard statements:
guard let xyz = xyz else {
// Handle failure and then exit this code block
// e.g. by calling return, break, continue, or throw
return
}
// Do something with xyz, which is now guaranteed to be non-nil
Note how ordinary optional binding can lead to greater indentation when there is more than one optional value:
if let abc = abc {
if let xyz = xyz {
// Do something with abc and xyz
}
}
You can avoid this nesting with guard statements:
guard let abc = abc else {
// Handle failure and then exit this code block
return
}
guard let xyz = xyz else {
// Handle failure and then exit this code block
return
}
// Do something with abc and xyz
Swift 5 Protocol Extension
Here is an approach using protocol extension so that you can easily inline an optional nil check:
import Foundation
public extension Optional {
var isNil: Bool {
guard case Optional.none = self else {
return false
}
return true
}
var isSome: Bool {
return !self.isNil
}
}
Usage
var myValue: String?
if myValue.isNil {
// do something
}
if myValue.isSome {
// do something
}
One option that hasn't specifically been covered is using Swift's ignored value syntax:
if let _ = xyz {
// something that should only happen if xyz is not nil
}
I like this since checking for nil feels out of place in a modern language like Swift. I think the reason it feels out of place is that nil is basically a sentinel value. We've done away with sentinels pretty much everywhere else in modern programming so nil feels like it should go too.
Instead of if, ternary operator might come handy when you want to get a value based on whether something is nil:
func f(x: String?) -> String {
return x == nil ? "empty" : "non-empty"
}
Another approach besides using if or guard statements to do the optional binding is to extend Optional with:
extension Optional {
func ifValue(_ valueHandler: (Wrapped) -> Void) {
switch self {
case .some(let wrapped): valueHandler(wrapped)
default: break
}
}
}
ifValue receives a closure and calls it with the value as an argument when the optional is not nil. It is used this way:
var helloString: String? = "Hello, World!"
helloString.ifValue {
print($0) // prints "Hello, World!"
}
helloString = nil
helloString.ifValue {
print($0) // This code never runs
}
You should probably use an if or guard however as those are the most conventional (thus familiar) approaches used by Swift programmers.
Optional
Also you can use Nil-Coalescing Operator
The nil-coalescing operator (a ?? b) unwraps an optional a if it contains a value, or returns a default value b if a is nil. The expression a is always of an optional type. The expression b must match the type that is stored inside a.
let value = optionalValue ?? defaultValue
If optionalValue is nil, it automatically assigns value to defaultValue
Now you can do in swift the following thing which allows you to regain a little bit of the objective-c if nil else
if textfieldDate.text?.isEmpty ?? true {
}
var xyz : NSDictionary?
// case 1:
xyz = ["1":"one"]
// case 2: (empty dictionary)
xyz = NSDictionary()
// case 3: do nothing
if xyz { NSLog("xyz is not nil.") }
else { NSLog("xyz is nil.") }
This test worked as expected in all cases.
BTW, you do not need the brackets ().
If you have conditional and would like to unwrap and compare, how about taking advantage of the short-circuit evaluation of compound boolean expression as in
if xyz != nil && xyz! == "some non-nil value" {
}
Granted, this is not as readable as some of the other suggested posts, but gets the job done and somewhat succinct than the other suggested solutions.
If someone is also try to find to work with dictionaries and try to work with Optional(nil).
let example : [Int:Double?] = [2: 0.5]
let test = example[0]
You will end up with the type Double??.
To continue on your code, just use coalescing to get around it.
let example : [Int:Double?] = [2: 0.5]
let test = example[0] ?? nil
Now you just have Double?
This is totally logical, but I searched the wrong thing, maybe it helps someone else.
Since Swift 5.7:
if let xyz {
// Do something using `xyz` (`xyz` is not optional here)
} else {
// `xyz` was nil
}