[Swift] Table View Cell, how to jump to an external URL

Introduction

I wanted to use TableViewCell and put a link to my homepage in the app, but I couldn't find it right away, so I'll write it as a reminder.

• Because it is a beginner, I would appreciate it if you could tell me if there are any mistakes.

Finished product

In this explanation, tapping "Google" in TabelViewCell will take you to the "Google" page.

• This time, I used Google's page as a sample.

Set Table View on Story board

Give the Identifier a name

This time, I chose "Cell".

Create a TableViewControllre.swif file

Connect TableView and TableViewController.swift

Prepare an array to put in the cell

OutisideTableViewController.swift

 //This time I will link to Google
    var outsideArray = ["Google"]

Put a value in a cell

OutisideTableViewController.swift

 override func numberOfSections(in tableView: UITableView) -> Int {
        // #warning Incomplete implementation, return the number of sections
        return 1
    }

    override func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        // #warning Incomplete implementation, return the number of rows
        return outsideArray.count
    }

    
    override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        //Put the name set in Identifier in sithIdentifier
        let cell = tableView.dequeueReusableCell(withIdentifier: "Cell", for: indexPath)

        cell.textLabel?.text = outsideArray[indexPath.row]

        return cell
    }

Process to open an external browser

OutisideTableViewController.swift

//Processing when a cell is tapped
    override func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
        
        //Erases the selected color when tapped
        tableView.deselectRow(at: indexPath as IndexPath, animated: true)
        
        //Open the URL in an external browser
        let url = NSURL(string: "https://www.google.com/?hl=ja")
        if UIApplication.shared.canOpenURL(url! as URL) {
            UIApplication.shared.open(url! as URL, options: [:], completionHandler: nil)
        }
    }

Full code

OutisideTableViewController.swift

import UIKit

class OutsideTableViewController: UITableViewController {
    
   //This time I will link to Google
    var outsideArray = ["Google"]

    override func viewDidLoad() {
        super.viewDidLoad()

    }

    override func numberOfSections(in tableView: UITableView) -> Int {
        // #warning Incomplete implementation, return the number of sections
        return 1
    }

    override func tableView(_ tableView: UITableView, numberOfRowsInSection section: Int) -> Int {
        // #warning Incomplete implementation, return the number of rows
        return outsideArray.count
    }

    
    override func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
        //Put the name set in Identifier in sithIdentifier
        let cell = tableView.dequeueReusableCell(withIdentifier: "Cell", for: indexPath)

        cell.textLabel?.text = outsideArray[indexPath.row]

        return cell
    }
    
    //Processing when a cell is tapped
    override func tableView(_ tableView: UITableView, didSelectRowAt indexPath: IndexPath) {
        
        //Erases the selected color when tapped
        tableView.deselectRow(at: indexPath as IndexPath, animated: true)
        
        //Open the URL in an external browser
        let url = NSURL(string: "https://www.google.com/?hl=ja")
        if UIApplication.shared.canOpenURL(url! as URL) {
            UIApplication.shared.open(url! as URL, options: [:], completionHandler: nil)
        }
    }
}

Reference site http://somen.site/2018/09/24/%E3%82%A2%E3%83%97%E3%83%AA%E3%81%8B%E3%82%89%E5%A4%96%E9%83%A8%E3%83%96%E3%83%A9%E3%82%A6%E3%82%B6%E3%81%A7url%E3%82%92%E9%96%8B%E3%81%8F-swift4-1/ https://pg-happy.jp/swift-tableview-tableviewcell.html