Swift 3 Open Link

Swift 3 Open Link

Use guard to unwrap the textfield text property, replacing the occurrences, add percent encoding to the result and create an URL from the resulting string:

Try like this:

guard
let text = birdName.text?.replacingOccurrences(of: " ", with: "+"),
let query = text.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed),
let url = URL(string: "https://google.com/#q=" + query)
else { return }
if #available(iOS 10.0, *) {
UIApplication.shared.open(url)
} else {
UIApplication.shared.openURL(url)
}

How to open another View from links within UIWebview in Swift 3?

From inside if block, you need to return false, and it should work.
By returning false the delegate method tells that the webView should not start loading the url.

public func webView(_ webView: UIWebView, shouldStartLoadWith request: URLRequest, navigationType: UIWebViewNavigationType) -> Bool {

if request.url?.scheme == "myapp" {
// do your work here
UIApplication.shared.open(request.url!, options: [:], completionHandler: nil)
//Add the below line of code to the existing code
return false
}
return true

}

How to open an URL in Swift?

All you need is:

guard let url = URL(string: "http://www.google.com") else {
return //be safe
}

if #available(iOS 10.0, *) {
UIApplication.shared.open(url, options: [:], completionHandler: nil)
} else {
UIApplication.shared.openURL(url)
}

How to open an URL with parameter in Swift3

Yea I figure out the problem path parameter not working anymore but query parameter working

func shareToInstagram() {

let instagramURL = URL(string: "instagram://user?username=mark20")

if (UIApplication.shared.canOpenURL(instagramURL! as URL)) {

UIApplication.shared.open(instagramURL!, options: ["":""], completionHandler: nil)

print("miss you so much ")

} else {
print(" Instagram isn't installed ")
}
}

more check this link instagram iPhone Hooks

Opening an url in the background with swift 3?

I believe what you're looking for is an HTTP request. This is what browsers do to get content from a webpage. There are many ways to do an HTTP request to a url in iOS. One such way is with the URLSession class.

let url = URL(string: "http://www.google.com/") //Or your URL
var request = URLRequest(url: url!)
request.httpMethod = "POST"
request.httpBody = "Data to send to your backend".data(using: .utf8)!

let task = URLSession.shared.dataTask(with: request) { data, response, error in
if error != nil {
//There was an error
} else {
//The HTTP request was successful
print(String(data: data!, encoding: .utf8)!)
}

}
task.resume()

This will send the data in the httpBody property to your backend in a HTTP request.



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