I have tried
self.adc_role_id = String(res["adc_role_id"])
self.adc_role_id = "\(res["adc_role_id']"
self.adc_role_id = (\(res["adc_role_id"] as? String)!
but still get
Could not cast value of type '__NSCFNumber' to 'NSString'
I added the dump of res[4] below
As new as I am to Swift, I don't know anything else to try
In Swift 4, the String initializer requires the describing: argument label.
I don't know if this will solve your problem, but your first line of code should be written:
self.adc_role_id = String(describing: res["adc_role_id"])
In your screenshot we can see that res["adc_role_id"] is an NSNumber.
To transform an NSNumber to a String you should use its stringValue property.
And since a dictionary gives an Optional, you should use optional binding to safely unwrap it.
Example:
if let val = res["adc_role_id"] {
self.adc_role_id = val.stringValue
}
You could also, if you want, use string interpolation instead of the property:
if let val = res["adc_role_id"] {
self.adc_role_id = "\(val)"
}
but I think using the property is more relevant.
If for some reason the compiler complains about the type of the content, cast it:
if let val = res["adc_role_id"] as? NSNumber {
self.adc_role_id = val.stringValue
}
Note that you should not use String(describing:) because this initializer will try to represent the string in many ways, and some of them will give inaccurate and unexpected results (for example, if String(describing:) resolves to use the debugDescription property, as explained in the documentation, you may get a totally different string than the one you want).
It's also worth noting that using String(describing:) with an optional value such as your dictionary will resolve to a wrong string: String(describing: res["adc_role_id"]) will give Optional(yourNumber)! This is why Mike's answer is wrong. Be careful about this. My advice is to avoid using String(describing:) altogether unless for debugging purposes.
The error message is clear and the dump is clear, too.
The value is not a String, it's an Int(64) wrapped in NSNumber
Optional bind the value directly to Int (NSNumber is implicit bridged to Int) and use the String initializer.
if let roleID = res["adc_role_id"] as? Int {
self.adc_role_id = String(roleID)
}
Please conform to the naming convention that variable names are camelCased rather than snake_cased
Related
I make a query that returns a NSNumber. I then attempt to cast the NSNumber to String. For some reason, it always prints/ compares as an optional...but when I check the variables type it says string...Why is it optional? I need it to be a string!
let whoseTurn = selectedGame?["currentSender"]
let whoseTurnAsString: String = String(describing: whoseTurn)
if let whoseTurn = selectedGame?["currentSender"] as? NSNumber {
let whoseTurnAsString = "\(whoseTurn)"
print(whoseTurnAsString)
}
This is the right way to do optional chaining and make sure you are not forcing an optional
whoseTurn is an optional wrapping your NSNumber. You aren't "casting" it here to a string, you are making a string that "describes" it, and in this case that description includes the fact that whoseTurn is an optional... If you don't want that you'll need to unwrap it,
let whoseTurnAsString: String = String(describing: whoseTurn!)
(note the ! at the end)
This line of code let whoseTurn = selectedGame?["currentSender"] will return an optional.
This line of code let whoseTurnAsString: String = String(describing: whoseTurn) will return a String describing that optional value, which will be a string like this: Optional(5) or Optional(6). It describes the value saying that it is an optional.
So you need to unwrap that optional in order to get the wrapped value, you can force unwrap selectedGame like this:
let whoseTurn = selectedGame!["currentSender"] and then use the normal String initializer like this: String(whoseTurn).
Or, preferably, safely unwrap it like this:
if let whoseTurn = selectedGame?["currentSender"] {
let whoseTurnAsString = String(whoseTurn)
}
String can be optional also '?' or '!' indicates optional, check documentation on optionals.
I just joined a project that has a lot of existing code. The previous programmer was perhaps unfamiliar with Swift or began development in the early stages of the Swift language. They seemed to be using the if let statement in an odd way. They seemed to want to use the statement as a if is let. Before I edit the code I would like to know if there is any valid use for this:
// In JSON parser
if value is String, let string = value as? String {
document.createdBy = string
}
First checking if value is of type String seems redundant to me. Doesn't Swift check for this in the let string = value as? String portion of the statement?
QUESTION
Why would this need to be checked twice? Or would there be a reason for this?
You're correct, this is redundant. If value is not a string, then value as? String would return nil, and the conditional binding would fail.
To check the type, and not use the casted result:
if value is String {
// Do something that doesn't require `value` as a string
}
To check the type and use the result:
if let value = value as? String { // The new name can shadow the old name
document.createdBy = value
}
Doing both makes no sense.
Say I have
var dict = parseJSON(getJSON(url)) // This results in an NSDictionary
Why is
let a = dict["list"]![1]! as NSDictionary
let b = a["temp"]!["min"]! as Float
allowed, and this:
let b = dict["list"]![1]!["temp"]!["min"]! as Float
results in an error:
Type 'String' does not conform to protocol 'NSCopying'
Please explain why this happens, note that I'm new to Swift and have no experience.
dict["list"]![1]! returns an object that is not known yet (AnyObject) and without the proper cast the compiler cannot know that the returned object is a dictionary
In your first example you properly cast the returned value to a dictionary and only then you can extract the value you expect.
To amend the answer from #giorashc: use explicit casting like
let b = (dict["list"]![1]! as NSDictionary)["temp"]!["min"]! as Float
But splitting it is better readable in those cases.
how can i convert any object type to a string?
let single_result = results[i]
var result = ""
result = single_result.valueForKey("Level")
now i get the error: could not assign a value of type any object to a value of type string.
and if i cast it:
result = single_result.valueForKey("Level") as! String
i get the error:
Could not cast value of type '__NSCFNumber' (0x103215cf0) to 'NSString' (0x1036a68e0).
How can i solve this issue?
You can't cast any random value to a string. A force cast (as!) will fail if the object can't be cast to a string.
If you know it will always contain an NSNumber then you need to add code that converts the NSNumber to a string. This code should work:
if let result_number = single_result.valueForKey("Level") as? NSNumber
{
let result_string = "\(result_number)"
}
If the object returned for the "Level" key can be different object types then you'll need to write more flexible code to deal with those other possible types.
Swift arrays and dictionaries are normally typed, which makes this kind of thing cleaner.
I'd say that #AirSpeedVelocity's answer (European or African?) is the best. Use the built-in toString function. It sounds like it works on ANY Swift type.
EDIT:
In Swift 3, the answer appears to have changed. Now, you want to use the String initializer
init(describing:)
Or, to use the code from the question:
result = single_result.valueForKey("Level")
let resultString = String(describing: result)
Note that usually you don't want valueForKey. That is a KVO method that will only work on NSObjects. Assuming single_result is a Dictionary, you probably want this syntax instead:
result = single_result["Level"]
This is the documentation for the String initializer provided here.
let s = String(describing: <AnyObject>)
Nothing else is needed. This works for a diverse range of objects.
The toString function accepts any type and will always produce a string.
If it’s a Swift type that implements the Printable protocol, or has overridden NSObject’s description property, you’ll get whatever the .description property returns. In the case of NSNumber, you’ll get a string representation of the number.
If it hasn’t, you’ll get a fairly unhelpful string of the class name plus the memory address. But most standard classes, including NSNumber, will produce something sensible.
import Foundation
class X: NSObject {
override var description: String {
return "Blah"
}
}
let x: AnyObject = X()
toString(x) // return "Blah"
"\(x)" // does the same thing but IMO is less clear
struct S: Printable {
var description: String {
return "asdf"
}
}
// doesn't matter if it's an Any or AnyObject
let s: Any = S()
toString(s) // reuturns "asdf"
let n = NSNumber(double: 123.45)
toString(n) // returns "123.45"
n.stringValue // also works, but is specific to NSNumber
(p.s. always use toString rather than testing for Printable. For one thing, String doesn’t conform to Printable...)
toString() doesn't seem to exist in Swift 3 anymore.
Looks like there's a failable initializer that will return the passed in value's description.
init?(_ description: String)
Docs here https://developer.apple.com/reference/swift/string/1540435-init
I have an "if let" statement that is being executed, despite the "let" part being nil.
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]! {
println(leftInc)
let valueString : String = formatter.stringFromNumber(NSNumber(double: leftInc!))!
self.leftIncisorTextField?.text = valueString
self.btnLeftIncisor.associatedLabel?.text = valueString
}
// self.analysis.inputs is a Dictionary<String, Double?>
The inputs dictionary holds information entered by the user - either a number, or nil if they haven't entered anything in the matching field yet.
Under the previous version of Swift, the code was written as this:
if let leftInc : Double? = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!]?? {
and worked correctly.
I saw a similar question here, but in that instance the problem seemed to be the result of using Any?, which is not the case here.
Swift 2.2
In your if let you define another optional, that's why nil is a legitimate case. if let is intended mainly to extract (maybe) non optional value from an optional.
You might try:
if let leftInc : Double = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!].flatMap ({$0}) {
// leftInc is not an optional in this scope
...
}
Anyway I'd consider to not do it as a one liner but take advantage of guard case. Just in order to enhance readability. And avoid bang operator (!).
The if-let is for unwrapping optionals. You are allowing nil values by setting the type to an optional Double.
The if statement should be:
if let leftInc = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!] as? Double{
...
}
This will attempt to get an object out of inputs, if that fails it returns nil and skips it. If it does return something it will attempt to convert it to a Double. If that fails it skips the if statement as well.
if inputs is a dictionary like [Something:Double] then you don't need the last as? Double as indexing the dictionary will return a Double?
I recommend reading the swift book on optional chaining.
You could break it down further -
if let optionalDouble = self.analysis.inputs[self.btnLeftIncisor.dictionaryKey!], leftInc = optionalDouble {
....
}
as your dictionary has optional values - this way of writing it might make it clearer what's going on
if let k = dict["someKey"]{}, dict["someKey"] will be an object of type Any
this can bypass a nill
So do a typecast to get it correct like if let k = dict["someKey"] as! String {}