I am asking this hesitantly as I know this is probably a dumb question.
I am returning a Realm result and then have gone ahead and tried to cast it to a String as normal (to put in a text label).
However I'm getting an error 'init' has been renamed to 'init(describing:)'.
When I try use the describing method instead, the label prints "Optional" inside it which obviously isn't what I want.
Is there a reason I can't use :
previousTimeLabel.text = String(lastRecord?.time)
I'm sure I've done this before and it's been fine, am I missing something? (lastRecord.time is an Int).
I've checked the answer here about Interpolation Swift String Interpolation displaying optional? and tried changing to something like this :
if let previousRounds = String(lastRecord?.rounds) {
previousRoundsLabel.text = previousRounds
}
but get the same error + Initializer for conditional binding must have Optional type, not 'String'
The issue isn't String(lastRecord?.time) being Optional. The issue is lastRecord being Optional, so you have to unwrap lastRecord, not the return value of String(lastRecord?.time).
if let lastRecord = lastRecord {
previousRoundsLabel.text = "\(lastRecord.time)"
}
To summarize Dávid Pásztor's answer, here's a way you can fix it:
previousTimeLabel.text = String(lastRecord?.time ?? 0)
This may not be the best way for your application. The point Dávid was making is that you need to deal with lastRecord possibly being nil before trying to pass its time Int into the String initializer. So the above is one simple way to do that, if you're ok with having "0" string as your previousTimeLabel's text if there was no lastRecord.
Related
I am declaring a constant in this line of code but if I don't put the `! after it, xXcode gives an error saying:
value of optional type string must be unwrapped.
Why does Xcode think this is an optional? I am just declaring a constant of type String and assigning it to a key that the user will set in the setting section.
Maybe because I am using the UserDefault settings and it's not set yet? If so, how do I get around that?
let jbEmail: String = userDefaults.string(forKey: "JBemail_preference")!
Look at the documentation for UserDefaults string(forKey:). It has a return type of String?. It returns an optional because there might not be a value for the given key.
So your attempt to assign a String? to a String results in the error. The forced unwrap (adding !) resolves the error but it is the worst possible solution because now your app will crash if there is no value for the key.
You should properly handle the situation where there is no value for the key in UserDefaults.
You can assign a default value:
let jbEmail = userDefaults.string(forKey: "JBemail_preference") ?? "Some Default"
Or you can conditional deal with there being no value:
if let jbEmail = userDefaults.string(forKey: "JBemail_preference") {
// Do something with jbEmail
} else {
// There is no value, do something else
}
I'm receiving a compiler error and I'm not really sure why. I'm sure there is a simple answer for this. I have a core data attribute I'm trying to assign before saving. In my Core Data Property file it's defined as this:
#NSManaged public var age: Int32
I am using a UIPicker to select it and put it into an inputView. That works fine, so ageTextField: UITextField! holds the value. As I try to assign this to the CoreData object just before saving I get the following
person.age = ageTextField.text -> Cannot assign String? to Int32.
Ok, I understand that, so I cast it
person.age = Int(ageTextField.text) -> Value of Optional String not unwrapped...
Ok, I get that, so I unwrapped it, it asks to unwrap again and I agree:
person.age = Int(ageTextField.text!)! -> Type of expression is ambiguous without more context
I'm not sure what is wrong here, just looking over some old Swift 2 code of mine and this worked. This is my first code with Swift 3 though.
That compiler error is obscure at best and misleading at worst. Change your cast to Int32:
person.age = Int32(ageTextField.text!)!
Also: unless you are absolutely sure that the user will always enter a valid number into the textfield, use optional binding instead of force unwrap:
if let text = ageTextField.text,
let age = Int32(text)
{
person.age = age
}
The immediate issue is the use of the wrong type. Use Int32, not Int. But even once that is fixed, you have lots of other issues.
You should safely unwrap the text and then the attempt to convert the string to an integer.
if let text = ageTextField.text, let num = Int32(text) {
person.age = num
}
The use of all of those ! will cause a crash if the text is nil or it contains a value that isn't a valid number. Always use safe unwrapping.
Just make sure to unwrap the optional before convert. Make your code safe.
I'm very new to swift, but proficient in other languages like Java, JavaScript, C, ... I'm lost with Swift syntax when it comes to create expressions. Look at this basic example where I just try to find out if one string is contained into another by calling String.rangeOfString that returns an Optional Range (Range?)
This works as expected:
let LEXEMA:String="http://"
let longUrl:String="http://badgirls.many/picture.png"
let range=longUrl.rangeOfString(LEXEMA);
if (range? != nil) {
// blah
}
Now I'm trying to combine the expression inside the if, something like:
if (longUrl.rangeOfString(LEXEMA)? !=nil) {
// blah
}
But I always get syntax errors, the above yields a "Expected Separator" and can't understand why. Done some more tests:
if (absolutePath.rangeOfString(URL_LEXEMA) !=nil) { }
Expected Separator before "!"
if absolutePath.rangeOfString(URL_LEXEMA) !=nil { }
Braced block of statements is an unused closure
What am I doing wrong?
If you’re coming from other like Java, you might be thinking of optionals like pointers/references, and so used to equating them to nil and if non-nil, using them. But this is probably going to lead to more confusion. Instead, think of them like a container for a possible result, that you need to unwrap to use. if let combines the test and unwrapping operation.
With this in mind, here’s how you could adapt your code:
let LEXEMA: String="http://"
let longUrl: String="http://badgirls.many/picture.png"
if let range = longUrl.rangeOfString(LEXEMA) {
// use range, which will be the unwrapped non-optional range
}
else {
// no such range, perhaps log an error if this shouldn’t happen
}
Note, that ? suffixing behaviour you were using changes in Swift 1.2 so even the code in your question that compiles in 1.1 won’t in 1.2.
It’s possible that sometimes you are whether there was a value returned, but you don’t actually need that value, just to know it wasn’t nil. In that case, you can compare the value to nil without the let:
if longUrl.rangeOfString(LEXEMA) != nil {
// there was a value, but you don't care what that value was
}
That said, the above is probably better expressed as:
if longUrl.hasPrefix(LEXEMA) { }
For starters:
You don't need parenthesis with if statements unless you have nested parenthetical subexpressions that require it.
You don't need to specify the type on the left side of the = of a let or var declaration if Swift can figure it out from the right side of the =. Very often Swift can figure it out, and you can tell that Swift can figure it out, so you can avoid that redundant clutter.
You do need to specify the type if Swift cannot figure out the type from
the right side. Example:
For example, consider the following lines:
let LEXEMA = "http://"
let longUrl = "http://badgirls.many/picture.png"
Swift can figure out that they're strings.
Similarly for this function or class that returns a UIView:
var myView = ViewReturningClassOrFunc()
Consider this:
#IBOutlet var myView : UIView!
In the above line, Swift cannot figure out ahead of time it will be assigned a UIView, so you have to provide the type. By providing a ! at the end you've made it an implicitly unwrapped optional. That means, like ?, you're indicating that it can be nil, but that you are confident it will never be nil at the time you access it, so Swift won't require you to put a ! after it when you reference it. That trick is a time saver and big convenience.
You should NOT add the ? to the line:
if (longUrl.rangeOfString(URL_LEXEMA) !=nil) {
As another answer pointed out, you're missing the let.
if let longUrl.rangeOfString(URL_LEXEMA) {
println("What do I win? :-)")
}
swift is case sensitive language. you need to check about whitespaces as well
if longUrl.rangeOfString(LEXEMA) != nil {
//your condition
}
there should be space between statement != nil
Just add a space between != and nil like:
if longUrl.rangeOfString(LEXEMA) != nil {
// blah
}
I tested your code in playground, an error of Expected ',' separator reported.
And do not forget the rules that 1s and 0s and Airspeed Velocity said.
I am reading the following book, on Page #32 there is a code snippet. December 2014: First Edition.
Swift Development with CocoaJonathon Manning, Paris Buttfield-Addison,
and Tim Nugent
I know we can use ? to make a vairbale optional and ! to unwrap a optional vairbale in Swift
var str: String? = "string"
if let theStr = str? {
println("\(theStr)")
} else {
println("nil")
}
Why do we need ? in this line if let theStr = str? It seems working just fine with out it and making no difference.
You don't need it, and shouldn't have it. The optional binding evaluates the optional. If it's nil, it stops. If it's not nil, it assigns the value to your required variable and then executes the code inside the braces of the if statement.
EDIT:
The language has changed slightly since it was first given out in beta form. My guess is that the ? used to be required.
Some of the sample projects I've used from Github fail to compile and I've had to edit them to get them to work. this might be an example where the syntax has changed slightly.
The current version of Swift does not require it, as it is redundant since your variable is already an optional.
Whatever you put in the if let statement does have to be an optional though or you will receive an error
Bound Value in a conditional binding must be of Optional type
Furthermore, if you are casting to a type, you do need to use as? to cast to an optional type.
var str2: Any = ["some", "example"]
if let theStr = str2 as? [String] {
println("\(theStr)")
} else {
println("nil")
}
I'm trying to call substringWithRange on an option String but after multiple experiments am still not able to get it to compile:
var mdn:String?
var subscriber = CTSubscriber()
var carrierToken = subscriber.carrierToken
mdn = NSString(data:carrierToken, encoding:NSUTF8StringEncoding)
let range:NSRange = NSRange(location: 0, length: 10)
if mdn
{
let subString = mdn!.substringWithRange(range)
}
This will result in the compilation error saying the value of the optional NSString is not unwrapped.
I thought it already was unwrapped due to the !.
If I remove ! then I get an error saying String? does not have a member named substringWithRange.
Replace
var mdn:String?
with
var mdn:NSString?
You are using the var to store a NSString so you should give it the correct type. Although String and NSString are mutually assignable, it's not the same type.
String doesn't have the substringWithRange method.
I think this might actually be a compiler bug. I would report it to Apple and see how they come back.
However, the proper approach to unwrapping an optional for use if you need to gain some information from it. (Like obtaining a substring, in your example) is:
if let theMdn = mdn {
let subString = theMdn.substringWithRange(range)
}
That way your codeblock will only be executed if the unwrapping was successful.
Edit: Sulthan's answer might be more accurate. However I still think if you can unwrap the optional in a let block, you should be able to do it manually.