String With Contents Of URL?

Question

So before I downloaded the recent update, the following code worked for me:

var g_home_url = String.stringWithContentsOfURL(NSURL(string: url_string), encoding: NSUTF8StringEncoding, error: nil) // Gives me an error: "String.Type does not have a member names stringWithContentsOfUrl"

I am confused. What is the proper way to acieve the following objective-c method in swift?

NSString * g_home_url = [[NSString alloc] initWithContentsOfURL:[NSURL URLWithString:home_url] encoding:NSUTF8StringEncoding error:nil];

Guillaume Algis · Accepted Answer · 2014-10-23T16:02:49.770

21

Use the -initWithContentsOfURL:encoding:error: instance method instead of the +stringWithContentsOfURL:encoding:error: class convenience initializer.

var g_home_url = String(contentsOfURL: NSURL(string: url_string)!, encoding: NSUTF8StringEncoding, error: nil)

I have no idea if class convenience initializers are now unsupported in Swift, but it would make sense as they were just shorthands for the alloc-init boilerplate, which doesn't exist in Swift.

edited Oct 23 '14 at 16:02

answered Oct 23 '14 at 15:56

Guillaume Algis

10,705
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1

`initWithContentsOfURL` and `stringWithContentsOfURL` are mapped to the *same* Swift method, and it seems that only the init method can be called from Swift in such a case. http://stackoverflow.com/questions/26491508/creating-an-object-in-swift-using-the-objective-c-factory-method-gives-a-compile describes a similar issue. – Martin R Oct 23 '14 at 16:11
`contentsOfURL` will fetch the response for that url. I don't know what's the reason. I use the quote syntax to convert it now. http://stackoverflow.com/questions/27062454/converting-url-to-string-and-back-again/31586194#31586194 – Roger Jul 23 '15 at 11:46

score 11 · Answer 2 · answered Dec 11 '16 at 09:07

11

For Swift 3 you'll have to use String(contentsOf:encoding:). It throws.

do {
    var content = try String(contentsOf:URL(string: "http://your-URI-here")!)
}
catch let error {
    // Error handling
}

answered Dec 11 '16 at 09:07

Niklas Berglund

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4
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Serhii Didanov · Answer 3 · 2018-06-07T12:34:14.030

At this moment (Swift 4.1.2) the only one way to send HTTP-get request from Ubuntu linux is: https://github.com/dmcyk/SwiftyCurl

import Foundation
import SwiftyCurl

var request = cURLRequest(url: URL(string: "https://path.to/url/")!, method: .get)
let connection = cURLConnection(useSSL: true)

do {
  let res = try connection.request(request)
  if let body: String = res.body() {
     print(body)
  }
} catch {
   prin(error)
}

score 0 · Answer 4 · edited Sep 01 '17 at 11:55

0

Here what it worked for me: (There is no more error argument)

var g_home_url = try! String(contentsOfURL: NSURL(string: url_string)!, encoding: NSUTF8StringEncoding)

or if you want to handle the error:

do {
  var g_home_url = try String(contentsOfURL: NSURL(string: url_string)!, encoding: NSUTF8StringEncoding)
}
catch {
   print(error)
}

edited Sep 01 '17 at 11:55

Abhishek Jain

4,557
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answered Jul 20 '16 at 00:39

Manu

629
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6

score 0 · Answer 5 · answered Apr 06 '18 at 18:34

0

For Swift 4 they have changed string encoding. So now it will be,

do {
  var g_home_url = try String(contentsOfURL: URL.init(string: url_string)!, encoding: String.Encoding.utf8)
}
catch {
   print(error)
}

answered Apr 06 '18 at 18:34

Sasi

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2
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44

`do { let ipAddress = try String(contentsOf: url, encoding: String.Encoding.utf8) } catch { print(error) }` – Ahmed Khedr Apr 27 '18 at 05:48
This solution don't on Linux with Swift 4.1.1. String(contentsOf: url) returns nothing – Serhii Didanov May 07 '18 at 12:26
@SergeyDi I encounter the same problem... I have also checked https://bugs.swift.org/browse/SR-7620 , but no update yet. Do you have any workaround? – 5t111111 May 10 '18 at 13:22
1

@5t111111 Only to use swift 3.x. But on 3.x version String(contentsOf also has an issue. Each time you call it, /tmp/someName.tmp file created. – Serhii Didanov May 11 '18 at 10:14

String With Contents Of URL?

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