CODEWITHSUNDEEP
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Sola
Sola

Reputation: 25

Random Number Generator in url

I currently have a random number generator in my url that is used to assign new numbers to the page parameter of the url. When I fetch a new image and print the url, it uses the same url. I do not know how to approach removing the existing random number in the url.

Here is my declaration:

public static var random = Int.random(in: 0..<450)
static let ImageURL = "https://api.unsplash.com/search/photos/?client_id=\(UnsplashClient.apiKey)&page=\(random))"

Upvotes: 0

Views: 786

Answers (3)

Mehdi Gilanpour
Mehdi Gilanpour

Reputation: 156

Do not use static for your array. Static variables only generated once. So it does not generate random numbers for you. read this: When to use static constant and variable in Swift?

Upvotes: 1

Frankenstein
Frankenstein

Reputation: 16361

Since you're using a static property and it's value is not gonna change each time you call it, unless you update it manually. You can fix this issue by using static computed property whose value is computed everytime it's used.

public static var random: Int { .random(in: 0..<450) }
public static var ImageURL: String { "https://api.unsplash.com/search/photos/?client_id=\(UnsplashClient.apiKey)&page=\(random))" }

Or you can combine them as @leo has mentioned in the comments, if you don't have any other purpose for random, like this:

public static var imageURL: String { "https://api.unsplash.com/search/photos/?client_id=\(UnsplashClient.apiKey)&page=\(Int.random(in: 0..<450)))" }

Upvotes: 0

iOS Developer
iOS Developer

Reputation: 470

According to me, You have to create a array of numbers that contains integers from 0 to 450 like this var array = Array(0...450) and and then can select random number from that array using array.randomElement()!. And after selecting, remove that number from that array.

Upvotes: 0

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