Swift Optional in Label

Swift optional in label

This is happening because the parameter you are passing to

String(stringInterpolationSegment:)

is an Optional.

Yes, you did a force unwrap and you still have an Optional...

Infact if you decompose your line...

let fundsreceived = String(stringInterpolationSegment: self.campaign?["CurrentFunds"]!)

into the following equivalent statement...

let value = self.campaign?["CurrentFunds"]! // value is an Optional, this is the origin of your problem
let fundsreceived = String(stringInterpolationSegment: value)

you find out that value is an Optional!

Why?

1. Because self.campaign? produces an Optional
2. Then ["CurrentFunds"] produces another Optional
3. Finally your force unwrap removes one Optional

So 2 Optionals - 1 Optional = 1 Optional

First the ugliest solution I can find

I am writing this solution just to tell you what you should NOT do.

let fundsreceived = String(stringInterpolationSegment: self.campaign!["CurrentFunds"]!)

As you can see I replaced your conditional unwrapping ? with a force unwrapping !. Just do not do it at home!

Now the good solution

Remember, you should avoid this guy ! everytime you can!

if let
    campaign = self.campaign,
    currentFunds = campaign["CurrentFunds"] {
        cell.FundsReceivedLabel.text = String(stringInterpolationSegment:currentFunds)
}

1. Here we are using conditional binding to transform the optional self.campaign into a non optional constant (when possible).
2. Then we are transforming the value of campaign["CurrentFunds"] into a non optional type (when possible).

Finally, if the IF does succeed, we can safely use currentFunds because it is not optional.

Hope this helps.

Unwrapping an Optional value for a Label

You are force unwrapping an optional by adding a !, so if the value is null like it is in your case the program will crash.

To unwrap an optional in a safe way (for a generic case) follow this method from Hacking With Swift:

var name: String? = nil
if let unwrapped = name {
print("\(unwrapped.count) letters")
} else {
print("Missing name.")
}

In your SPECIFIC case however verifiedLabel is likely your label set on a storyboard. So you can unwrap it like var verifiedLabel:UILabel! since the value should NOT be null.

So now since its null, a) please check your outlet connections and if everything looks good b) check this thread for debugging: IBOutlet is nil, but it is connected in storyboard, Swift

UIlabel text does not show optional word when setting optional text?

The text property of UILabel is optional. UILabel is smart enough to check if the text property's value is set to nil or a non-nil value. If it's not nil, then it shows the properly unwrapped (and now non-optional) value.

Internally, I imagine the drawRect method of UILabel has code along the lines of the following:

if let str = self.text {
    // render the non-optional string value in "str"
} else {
    // show an empty label
}

How do I remove optional text from UILabel

I would recommend Nil-Coalescing Operator ( ?? ). You can find detailed info here

class Test {
var ibu: Double?
}

var tt: Test? = Test()
tt?.ibu = 3.14
print(String(describing: tt?.ibu)) // Optional("3.14")
print(tt?.ibu ?? 0) // 3.14
tt?.ibu = nil
print(String(describing: tt?.ibu)) // nil
print(tt?.ibu ?? 0) // 0

Why is my label displaying Your cat is Optional(35) years old when it should display Your cat is 35 years old?

The problem is here:

String(describing: catYears)

catYears is an Optional<Int>, a string that describes an Optional<Int> will be in the format of Optional(<value>) or nil. That's why you get Optional(35).

You need to unwrap catYears!

String(describing: catYears!)

Or, remove String(describing:) all together and do:

if let humanYearsText = getHumanYears.text, let humanYears = Int(humanYearsText)
{
    let catYears = humanYears * 7
    displayCatYears.text = "Your cat is \(catYears) years old"
}

Showing optional double number on a label

Use if let to unwrap optional values:

let row = elementArray[indexPath.row]
if let myNumber = row.meltPoint {
    mPoint.text = "\(myNumber)"
} else {
     mPoint.text = "N/A"
}

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