How to get the First Character in a name - swift4

I have below func in my class.
static func getFirstCharInName(strName: String) -> String {
let firstCharInName = String(strName.first)
return firstCharInName.trim()
}
I encountered this err:
Value of optional type 'Character?' must be unwrapped to a value of type 'Character'
What seems to be the problem?
Thanks

func getFirstCharInName(strName: String) -> String {
let indexStartOfText = strName.index(strName.startIndex, offsetBy: 0)
let indexEndOfText = strName.index(strName.startIndex, offsetBy: 0)
let firstChar = String(strName[indexStartOfText...indexEndOfText])
return firstChar
}

This error means that the expression has optional value (the value can be nil) that is not yet unwrapped, strName.first returns an optional value of Character?, but your function demands a returning type of String which is not an optional type.
So, in order to fix this, you need to unwrap the optional value strName.first, it seems like you are not familiar with optionals, here's the code for your case (choose one from two options):
func getFirstCharInName(strName: String) -> String {
// option 1: force unwrap - can cause fatal error
return String(strName.first!)
// option 2: optional binding
if let firstCharInName = strName.first {
return String(firstCharInName)
} else {
// if the optional value is nil, return an empty string
return ""
}
}
PS. I don't really understand the function trim() in your question, but if you mean to strip away the blank spaces like " ", you can do:
firstCharInName.trimmingCharacters(in: .whitespaces)

Avoid the optional simply with prefix, it's totally safe. if there is no first character you'll get an empty string.
static func getFirstChar(in name: String) -> String { // the function name getFirstChar(in name is swiftier
return String(name.prefix(1))
}
I don't know what the trim function is supposed to do.

It means that value of optional type 'Character?' (as result of your part of code strName.first) must be unwrapped to a value of type 'Character' before you will be gonna cast it to String type.
You may use this variant:
func getFirstCharInName(strName: String) -> String {
return strName.count != 0 ? String(strName.first!) : ""
}
As you can see, the exclamation point is in the string strName.first! retrieves the optional variable as it was needed.

you can do something like that:
extension String {
var firstLetter: String {
guard !self.isEmpty else { return "" }
return String(self[self.startIndex...self.startIndex])
}
}
then
let name = "MilkBottle"
let first = name.firstLetter // "M"

Related

Swift Convert Optional String to Int or Int32 (Unwrapping optionals question)

I am trying to read a string and convert it to an int. I have a solution but it seems way too complicated. I guess I am still trying to wrap my head around unwrapping.
I have posted code below along with the compiler errors that I get with each solution.
In this example I try to read a string from UserDefaults and convert to an integer value.
static func GetSelectedSessionNum() -> Int32 {
var sessionNum : Int32 = 0
let defaults = UserDefaults.standard
let optionalString: String? = defaults.string(forKey: "selectedSessionNum")
// this works but it's too complicated
if let string = optionalString, let myInt = Int32(string) {
return myInt
}
return 0
// Error : optional String? must be unwrapped to a value of type 'String'
let t : String = defaults.string(forKey: "selectedSessionNum")
if let s : String = defaults.string(forKey: "selectedSessionNum") {
// error - Int32? must be unwrapped to a value of Int32
return Int32(s)
}
return 0
}
You need to cast to non optional Int32 in order to match your return type.
You can use any optional binding approach, or change your return type to Int32?
If you want an uncomplicated solution save selectedSessionNum as Int
static func getSelectedSessionNum() -> Int32 {
return Int32(UserDefaults.standard.integer(forKey: "selectedSessionNum"))
}
otherwise double optional binding
if let string = UserDefaults.standard.string(forKey: "selectedSessionNum"), let myInt = Int32(string) {
return myInt
}
or the nil coalescing operator
if let string = UserDefaults.standard.string(forKey: "selectedSessionNum") {
return Int32(string) ?? 0
}
is the proper way
If you want to avoid optional bindings, you can use flatMap, when called on Optional's it allows you to convert one optional to another:
return UserDefaults.standard.string(forKey: "selectedSessionNum").flatMap(Int32.init) ?? 0
You'd also need the ?? (nil coalescing operator) to cover the scenarios where either the initializer fails, or the value is not present in user defaults.

Swift string from optional Double

Is there a shortcut to specify placeholder text when the value is nil in Swift?
Right now I do:
let myText:String!
if myDouble != nil{
myText = "\(myDouble!)"
}else{
myText = "Value not provided"
}
That works, but it's very annoying to have to do that all the time. Is there a way to do something like
let myText:String = "\(myDouble ?? "Value no provided")"
That fails because it wants a default Double value, but I really want a String value.
You can use map and nil-coalescing:
let myText = myDouble.map { String($0) } ?? "Value not provided"
If myDouble is nil, the result of map is nil and the result is the value after the ??.
If myDouble is not nil, the result is the output of the map which creates a string from the Double.
For more details, please see the documentation for the map function of the Optional enumeration in the Swift standard library.
It seems reasonable to do
let myDouble: Double? = Double(3.0)
let myText = myDouble?.description ?? "Value not provided"
If myDouble is nil, then myText is "Value not provided". If myDouble is not nil, it's assigned the string representation of the number.
I think good approach to this is to make Extension to optional where Double is the Wrapped element:
extension Optional where Wrapped == Double {
var stringValue: String {
guard let me = self else { return "No Value Provided" }
return "\(me)"
}
}
// Use it like this:
myDouble.stringValue
Another approach could be making your custom operator like this:
public func ??(rhd: Double?, lhd: String) -> String {
if let unwrapped = rhd {
return String(unwrapped)
} else {
return lhd
}
}
And now your line let myText:String = "\(myDouble ?? "Value no provided")" works.
Please let me now if you don' understand anything.
This should work:
let myText = myDouble != nil ? String(myDouble!) : "Value not provided"

know the Datatype in Swift

i am new to swift i just started with the basics. In one of the Blog i saw a simple task which goes like this read a line from the stdin and check whether it is a integer,float,String.
I tried with the following code
let input = readLine()
var result = test(input)
print (result)
func test (obj:Any) -> String {
if obj is Int { return "This input is of type Intger." }
else if obj is String { return "This input is of type String." }
else { return "This input is something else. " }
}
when the input of 3245 is given it stores in the string format. and returns output as string.
how to overcome it..?
The readLine function returns a value of type String?. So your input variable can only be a String. It will never be Int or anything else.
If you want to see if the entered value is a valid number, you can try to convert the string to an Int.
if let input = readLine() {
if let num = Int(input) {
// the user entered a valid integer
} else {
// the user entered something other than an integer
}
}
As others have pointed out, readline() always returns a String?. It's up to you to parse that into whatever format you use it.
This is how I would do this:
let line = readLine()
switch line {
case let s? where Int(s) != nil:
print("This input is of type Intger.")
case let s? where Float(s) != nil:
print("This input is of type Float.")
case let s? where s.hasPrefix("\"") && s.hasSuffix("\""):
print("This input is of type String.")
default: print("This input is something else. ")
}
It exploits the ability of Int and Float's initializers to test the validity of a String, which almost entirely defeats the purpose of this exercise. But hey, it works, right? 😄
You can find of the type of object as
if let intt = obj as? Int {
// obj is a String. Do something with intt
}
else if let str = obj as? String {
// obj is a String. Do something with str
}
else {
//obj is something else
}

Function throws AND returns optional.. possible to conditionally unwrap in one line?

I am using an SQLite library in which queries return optional values as well as can throw errors. I would like to conditionally unwrap the value, or receive nil if it returns an error. I'm not totally sure how to word this, this code will explain, this is what it looks like:
func getSomething() throws -> Value? {
//example function from library, returns optional or throws errors
}
func myFunctionToGetSpecificDate() -> Date? {
if let specificValue = db!.getSomething() {
let returnedValue = specificValue!
// it says I need to force unwrap specificValue,
// shouldn't it be unwrapped already?
let specificDate = Date.init(timeIntervalSinceReferenceDate: TimeInterval(returnedValue))
return time
} else {
return nil
}
}
Is there a way to avoid having to force unwrap there? Prior to updating to Swift3, I wasn't forced to force unwrap here.
The following is the actual code. Just trying to get the latest timestamp from all entries:
func getLastDateWithData() -> Date? {
if let max = try? db!.scalar(eventTable.select(timestamp.max)){
let time = Date.init(timeIntervalSinceReferenceDate: TimeInterval(max!))
// will max ever be nil here? I don't want to force unwrap!
return time
} else {
return nil
}
}
Update: As of Swift 5, try? applied to an optional expression does not add another level of optionality, so that a “simple” optional binding is sufficient. It succeeds if the function did not throw an error and did not return nil. val is then bound to the unwrapped result:
if let val = try? getSomething() {
// ...
}
(Previous answer for Swift ≤ 4:) If a function throws and returns an optional
func getSomething() throws -> Value? { ... }
then try? getSomething() returns a "double optional" of the
type Value?? and you have to unwrap twice:
if let optval = try? getSomething(), let val = optval {
}
Here the first binding let optval = ... succeeds if the function did
not throw, and the second binding let val = optval succeeds
if the return value is not nil.
This can be shortened with case let pattern matching to
if case let val?? = try? getSomething() {
}
where val?? is a shortcut for .some(.some(val)).
I like Martin's answer but wanted to show another option:
if let value = (try? getSomething()) ?? nil {
}
This has the advantage of working outside of if, guard, or switch statements. The type specifier Any? isn't necessary but just included to show that it returns an optional:
let value: Any? = (try? getSomething()) ?? nil

Check empty string in Swift?

In Objective C, one could do the following to check for strings:
if ([myString isEqualToString:#""]) {
NSLog(#"myString IS empty!");
} else {
NSLog(#"myString IS NOT empty, it is: %#", myString);
}
How does one detect empty strings in Swift?
There is now the built in ability to detect empty string with .isEmpty:
if emptyString.isEmpty {
print("Nothing to see here")
}
Apple Pre-release documentation: "Strings and Characters".
A concise way to check if the string is nil or empty would be:
var myString: String? = nil
if (myString ?? "").isEmpty {
print("String is nil or empty")
}
I am completely rewriting my answer (again). This time it is because I have become a fan of the guard statement and early return. It makes for much cleaner code.
Non-Optional String
Check for zero length.
let myString: String = ""
if myString.isEmpty {
print("String is empty.")
return // or break, continue, throw
}
// myString is not empty (if this point is reached)
print(myString)
If the if statement passes, then you can safely use the string knowing that it isn't empty. If it is empty then the function will return early and nothing after it matters.
Optional String
Check for nil or zero length.
let myOptionalString: String? = nil
guard let myString = myOptionalString, !myString.isEmpty else {
print("String is nil or empty.")
return // or break, continue, throw
}
// myString is neither nil nor empty (if this point is reached)
print(myString)
This unwraps the optional and checks that it isn't empty at the same time. After passing the guard statement, you can safely use your unwrapped nonempty string.
In Xcode 11.3 swift 5.2 and later
Use
var isEmpty: Bool { get }
Example
let lang = "Swift 5"
if lang.isEmpty {
print("Empty string")
}
If you want to ignore white spaces
if lang.trimmingCharacters(in: .whitespaces).isEmpty {
print("Empty string")
}
Here is how I check if string is blank. By 'blank' I mean a string that is either empty or contains only space/newline characters.
struct MyString {
static func blank(text: String) -> Bool {
let trimmed = text.trimmingCharacters(in: CharacterSet.whitespacesAndNewlines)
return trimmed.isEmpty
}
}
How to use:
MyString.blank(" ") // true
You can also use an optional extension so you don't have to worry about unwrapping or using == true:
extension String {
var isBlank: Bool {
return self.trimmingCharacters(in: .whitespacesAndNewlines).isEmpty
}
}
extension Optional where Wrapped == String {
var isBlank: Bool {
if let unwrapped = self {
return unwrapped.isBlank
} else {
return true
}
}
}
Note: when calling this on an optional, make sure not to use ? or else it will still require unwrapping.
To do the nil check and length simultaneously
Swift 2.0 and iOS 9 onwards you could use
if(yourString?.characters.count > 0){}
isEmpty will do as you think it will, if string == "", it'll return true.
Some of the other answers point to a situation where you have an optional string.
PLEASE use Optional Chaining!!!!
If the string is not nil, isEmpty will be used, otherwise it will not.
Below, the optionalString will NOT be set because the string is nil
let optionalString: String? = nil
if optionalString?.isEmpty == true {
optionalString = "Lorem ipsum dolor sit amet"
}
Obviously you wouldn't use the above code. The gains come from JSON parsing or other such situations where you either have a value or not. This guarantees code will be run if there is a value.
Check check for only spaces and newlines characters in text
extension String
{
var isBlank:Bool {
return self.stringByTrimmingCharactersInSet(NSCharacterSet.whitespaceAndNewlineCharacterSet()).isEmpty
}
}
using
if text.isBlank
{
//text is blank do smth
}
Swift String (isEmpty vs count)
You should use .isEmpty instead of .count
.isEmpty Complexity = O(1)
.count Complexity = O(n)
isEmpty does not use .count under the hood, it compares start and end indexes startIndex == endIndex
Official doc Collection.count
Complexity: O(1) if the collection conforms to RandomAccessCollection; otherwise, O(n), where n is the length of the collection.
Single character can be represented by many combinations of Unicode scalar values(different memory footprint), that is why to calculate count we should iterate all Unicode scalar values
String = alex
String = \u{61}\u{6c}\u{65}\u{78}
[Char] = [a, l, e, x]
Unicode text = alex
Unicode scalar values(UTF-32) = u+00000061u+0000006cu+00000065u+00000078
1 Character == 1 extended grapheme cluster == set of Unicode scalar values
Example
//Char á == extended grapheme cluster of Unicode scalar values \u{E1}
//Char á == extended grapheme cluster of Unicode scalar values \u{61}\u{301}
let a1: String = "\u{E1}" // Unicode text = á, UTF-16 = \u00e1, UTF-32 = u+000000e1
print("count:\(a1.count)") //count:1
// Unicode text = a, UTF-16 = \u0061, UTF-32 = u+00000061
// Unicode text = ́, UTF-16 = \u0301, UTF-32 = u+00000301
let a2: String = "\u{61}\u{301}" // Unicode text = á, UTF-16 = \u0061\u0301, UTF-32 = u+00000061u+00000301
print("count:\(a2.count)") //count:1
For optional Strings how about:
if let string = string where !string.isEmpty
{
print(string)
}
if myString?.startIndex != myString?.endIndex {}
I can recommend add small extension to String or Array that looks like
extension Collection {
public var isNotEmpty: Bool {
return !self.isEmpty
}
}
With it you can write code that is easier to read.
Compare this two lines
if !someObject.someParam.someSubParam.someString.isEmpty {}
and
if someObject.someParam.someSubParam.someString.isNotEmpty {}
It is easy to miss ! sign in the beginning of fist line.
public extension Swift.Optional {
func nonEmptyValue<T>(fallback: T) -> T {
if let stringValue = self as? String, stringValue.isEmpty {
return fallback
}
if let value = self as? T {
return value
} else {
return fallback
}
}
}
What about
if let notEmptyString = optionalString where !notEmptyString.isEmpty {
// do something with emptyString
NSLog("Non-empty string is %#", notEmptyString)
} else {
// empty or nil string
NSLog("Empty or nil string")
}
You can use this extension:
extension String {
static func isNilOrEmpty(string: String?) -> Bool {
guard let value = string else { return true }
return value.trimmingCharacters(in: .whitespaces).isEmpty
}
}
and then use it like this:
let isMyStringEmptyOrNil = String.isNilOrEmpty(string: myString)