I'd like to append an s to the end of a string, if the value of a variable is > 1.
There are obviously several ways to accomplish this - for example:
if(points == 1) {
points_as_string = "1 Point"
} else {
points_as_string = "\(points) Points"
}
A shorter version could be:
points_as_string = (points == 1 ? "1 Point" : "\(points) Points")
An even shorter version would be:
points_as_string = "\(points) Point" + (points == 1 ? "" : "s")
Is it possible to write something even shorter than that, or is that as good as it gets?
Only very slightly shorter:
points_as_string = "\(points) Point\(points == 1 ? "" : "s")"
Here is a shorter version that uses a dictionary lookup and the nil coalescing operator ??:
points_as_string = "\(points) Point\([1:""][points] ?? "s")"
Related
swift: about ternary operator Question. Why my code is error code??? Please tell me why I'm wrong.
var arr = [0,1,2,3,4,5,6,7,8]
var result = 0;
for a in 0..<arr.count{
for b in 1..<arr.count - 1{
for c in 2..<arr.count - 2 {
arr[a] + arr[b] + arr[c] <= input[1] ? result = arr[a] + arr[b] +arr[c] : continue
}
}
}
[this is my error]
[1]: https://i.stack.imgur.com/UdiUB.png
In Swift, the ternary condition operator is an expression which takes the form
<condition> ? <expression if true> : <expression if false>
Expressions are part of larger statements, and the ternary specifically is one which evaluates to either the expression after the ?, or the one after the : depending on the truth of the condition.
continue, however, is not an expression but a statement on its own, which means that it cannot be on either side of the ternary.
Thinking about this another way: expressions evaluate to some value (e.g., can be put on the right-hand-side of an assignment, like x = <some expression>), while statements do not (e.g., it doesn't make sense to write x = continue).
You will need to express this in the form of a regular if-statement then:
if arr[a] + arr[b] + arr[c] <= input[1] {
result = arr[a] + arr[b] +arr[c]
} else {
continue
}
Note that the above code might be grammatically correct (in that it will compile), but it is unlikely to be what you mean: the loop will automatically continue at the end of execution even if arr[a] + arr[b] + arr[c] <= input[1] by default, which means that your result may get overwritten later in the loop. It seems likely that you mean something like
outer_loop: for a in 0 ..< arr.count {
for b in 1 ..< arr.count - 1 {
for c in 2 ..< arr.count - 2 {
if arr[a] + arr[b] + arr[c] <= input[1] {
result = arr[a] + arr[b] + arr[c]
// `break` would only exit the `c` loop, but with this label
// we can exit all loops at once.
break outer_loop
}
}
}
}
I'm using ag grid with angularjs and the filter does not work with formatted numbers. I use formatted numbers with currency values.
Below is the columndef code:
{ headerName:"GBO", field: "GBO", width: 200, editable:true, cellClass: "number-cell",filter:'agNumberColumnFilter',
cellRenderer : function(params){
if(params.value == "" || params.value == null)
return '-';
else return params.value;
}
}
Before assigning the data to the grid, I format the numbers using :
$scope.formatNumberOnly = function(num,c, d, t){
//console.log(num );
var n = getNumber(num);
//var n = this,
c = isNaN(c = Math.abs(c)) ? 2 : c,
d = d == undefined ? "." : d,
t = t == undefined ? "," : t,
s = n < 0 ? "-" : "",
i = parseInt(n = Math.abs(+n || 0).toFixed(c)) + "",
j = (j = i.length) > 3 ? j % 3 : 0;
return s + (j ? i.substr(0, j) + t : "") + i.substr(j).replace(/(\d{3})(?=\d)/g, "$1" + t) + (c ? d + Math.abs(n - i).toFixed(c).slice(2) : "");
};
});
The problem here is that the filter doesn't work with these formatted numbers and only seems to be working for values upto 999.
Can anyone please help me with a solution to this filtering problem?
If you want the filter to work on these formatted values, you should use a valueGetter instead of a valueFormatter
You should implement the above formatter function as a valueGetter in column Definition.
Also a number filter won't work as in order for your formatted number to be interpreted, it should be a text filter.
Here is an example from official docs.
After hours of Googling, I'm still at a standstill. I would appreciate if someone would point out the error in my formula or coding choice. Please keep in mind I'm new to Swift. I'm not used to non C-style for loops.
if textField.text != "" {
input = Double(textField.text!)! // parse input
// return if number less than 2 entered
if input < 2 {
resultLabel.text = "Enter a number greater than or equal to 2."
return;
}
// get square root of input and parse to int
inputSquared = Int(sqrt(input));
// loop from 2 to input iterating by 1
for i in stride(from: 2, through: input, by: 1) {
if inputSquared % Int(i) == 0 {
resultLabel.text = "\(Int(input)) is not a prime number."
}
else {
resultLabel.text = "\(Int(input)) is a prime number!"
}
}
}
I didn't know the formula on how to find a prime number. After looking up multiple formulas I have sorta settled on this one. Every result is a prime number, however. So my if condition is wrong. I just don't know how to fix it.
Check my algorithm.It works.But I'm not sure this is an effective algorithm for prime number
var input:Int = 30
var isPrime:Bool = true
if(input == 2){
print("Input value 2 is prim number")
}
else if(input < 2){
print("Input value must greater than 2")
}
else{
for i in 2...input-1{
if((input%i) == 0){
isPrime = false
break;
}
}
if(isPrime){
print("Your Input Value \(input) is Prime!")
}
}
A couple of solutions that work have been posted, but none of them explain why yours doesn't. Some of the comments get close, however.
Your basic problem is that you take the square root of input, then iterate from 2 to the input checking if the integer part of the square root is divisible by i. You got that the wrong way round. You need to iterate from 2 to the square root and check that the input is divisible by i. If it is, you stop because input is not prime. If you get to the end without finding a divisor, you have a prime.
try this code in playground you will get this better idea and try to use playground when you try the swift as you are not familiar with swift playground is best.
let input = 13 // add your code that take value from textfield
var prime = 1
// throw error less than 2 entered
if input < 2 {
assertionFailure("number should be 2 or greater")
}
// loop from 2 to input iterating by 1
for i in stride(from: 2, to: input, by: 1) {
if input % i == 0{
prime = 0
}
}
if prime == 1 {
print("\(input) number is prime")
} else {
print("\(input) number is not prime")
}
This code passes 3 as an argument to the arc4random_uniform() function and I guess that it returns true or false. Next you assign an enumerator to a variable. I don't understand what the function does, though.
let randomState = arc4random_uniform(3) == 2 ? CellState.Alive :
CellState.Empty
let cell = Cell(grid: self,
pos: (i, j),
state: randomState)
I am not following that logic.
Is it the "arc4random_uniform(3) == 2 ? CellState.Alive : CellState.Empty" format that you don't understand? It's written as a conditional (ternary) operator:
condition ? expr1 : expr2
Basically, if func arc4random_uniform(3) == 2, let randomState == CellState.Alive, otherwise let randomState == CellState.Empty.
Does anyone know if there is a way to use some kind shorthand in swift? more specifically, leaving out the braces in things like IF statements... eg
if num == 0
// Do something
instead of
if num == 0
{
// Do something
}
Those braces become rather space consuming when you have a few nested IF's.
PS. I do know I can do the following:
if num == 0 {
// Do something }
But I'm still curious if that sort of thing is possible
You can do that :
let x = 10, y = 20;
let max = (x < y) ? y : x ; // So max = 20
And so much interesting things :
let max = (x < y) ? "y is greater than x" : "x is greater than y" // max = "y is greater than x"
let max = (x < y) ? true : false // max = true
let max = (x > y) ? func() : anotherFunc() // max = anotherFunc()
(x < y) ? func() : anotherFunc() // code is running func()
This following stack : http://codereview.stackexchange.com can be better for your question ;)
Edit : ternary operators and compilation
By doing nothing more than replacing the ternary operator with an if else statement, the build time was reduced by 92.9%.
https://medium.com/#RobertGummesson/regarding-swift-build-time-optimizations-fc92cdd91e31#.42uncapwc
In swift you have to add braces even if there is just one statement in if:
if num == 0 {
// Do something
}
You cannot leave the braces, that how swift if statement work.
You could use a shorthand if statement like you would in objective-c:
num1 < num2 ? DO SOMETHING IF TRUE : DO SOMETHING IF FALSE
Swift 2.0 update
Method 1:
a != nil ? a! : b
Method 2: Shorthand if
b = a ?? ""
Referance: Apple Docs: Ternary Conditional Operator
and it does work,
u.dob = (userInfo["dob"] as? String) != nil ? (userInfo["dob"] as! String):""
I am replacing a json string with blank string if it is nil.
Edit: Adding Gerardo Medina`s suggestion...we can always use shorthand If
u.dob = userInfo["dob"] as? String ?? ""
It is called shorthand if-else condition. If you are into iOS development in Swift, then you can also manipulate your UI objects' behaviour with this property.
For e.g. - I want my button to be enabled only when there is some text in the textfield. In other words, should stay disabled when character count in textfield is zero.
button.enabled = (textField.characters.count > 0) ? true : false
its very simple :
in Swift 4
playButton.currentTitle == "Play" ? startPlay() : stopPlay()
Original Code is
if playButton.currentTitle == "Play"{
StartPlay()
}else{
StopPlay()
}
You could always put the entire if on one line:
if num == 0 { temp = 0 }