How can I add a link for a rate button with swift?

Swift 4

let url = URL(string: "itms-apps:itunes.apple.com/us/app/apple-store/id\(YOURAPPID)?mt=8&action=write-review")!
UIApplication.shared.openURL(url)

Try This, change appId in your method by your App ID

Swift 5

import StoreKit

func rateApp() {
    if #available(iOS 10.3, *) {
        SKStoreReviewController.requestReview()

    } else if let url = URL(string: "itms-apps://itunes.apple.com/app/" + "appId") {
        if #available(iOS 10, *) {
            UIApplication.shared.open(url, options: [:], completionHandler: nil)

        } else {
            UIApplication.shared.openURL(url)
        }
    }
}

Swift 3 \ 4

func rateApp() {
    guard let url = URL(string: "itms-apps://itunes.apple.com/app/" + "appId") else {
        return
    }
    if #available(iOS 10, *) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)

    } else {
        UIApplication.shared.openURL(url)
    }
}

id959379869 This is the id when you go on your Itune page of your app

Example : https://itunes.apple.com/fr/app/hipster-moustache/id959379869?mt=8

How get the ID :

  1. Itunesconnect account
  2. My Apps
  3. Click on "+" Button
  4. New iOS App
  5. Fill require details
  6. After filling all details goto your App
  7. Click on More Button
  8. View on AppStore
  9. It will redirect you to your App URL this will be universal
  10. Look Http URL

This is working the best for me. Directs the user straight to the 'Write A Review' composer of the application.

Swift 3.1 (Support for iOS10 and below)

Introduces new action=write-review

let appID = "959379869"

if let checkURL = URL(string: "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?id=\(appID)&pageNumber=0&sortOrdering=2&type=Purple+Software&mt=8") {
    open(url: checkURL)
} else {
    print("invalid url")
}

...

func open(url: URL) {
    if #available(iOS 10, *) {
        UIApplication.shared.open(url, options: [:], completionHandler: { (success) in
            print("Open \(url): \(success)")
        })
    } else if UIApplication.shared.openURL(url) {
            print("Open \(url)")
    }
}

Tested and works on Swift 2.2

let appID = "959379869" // Your AppID
if let checkURL = NSURL(string: "http://itunes.apple.com/WebObjects/MZStore.woa/wa/viewContentsUserReviews?id=\(appID)&pageNumber=0&sortOrdering=2&type=Purple+Software&mt=8") {
    if UIApplication.sharedApplication().openURL(checkURL) {
        print("url successfully opened")
    }
} else {
    print("invalid url")
}