Is it possible to have a viewController in an iOS app with some textboxes where the user inputs some strings and then with a button sends it to a webpage which is NOT mine? Here is an example of a webpage (http://www.nattklubbvictoria.se/#/list) with some choices, text boxes and a Send button. The websites is using some sort of widget. If it is possible, how?
Currently I have a uiweview but it's not so good looking and it's slow to load it. So it would be awesome if I could do this!
You could do something as simple as construct a GET URL & forward the user to that URL when the user clicks the button.
For example (not tested):
NSString *stringToSendUserTo = #"http://www.site.com/form.php?foreignParameterA=";
stringToSendUserTo = [stringToSendUserTo stringByAppendingString:firstTextField.text];
stringToSendUserTo = [stringToSendUserTo stringByAppendingString:#"&foreignParameterB="];
stringToSendUserTo = [stringToSendUserTo stringByAppendingString:secondTextField.text];
stringToSendUserTo = [stringToSendUserTo stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *URLToSendUserTo = [NSURL URLWithString:stringToSendUserTo];
if ([[UIApplication sharedApplication] canOpenURL:URLToSendUserTo]) {
[[UIApplication sharedApplication] openURL:URLToSendUserTo];
} else {
// do something
}
Related
Im trying to Get Information About a Website like title (Like what you see in safari in the Tabs and Windows)and Description of that Website from its URL Without Using the WebKit Framework in Xcode and Displaying it on screen as text. I Just want to know have to obtain the title of the Website/Page that the URL Belongs to using Cocoa. Is There a Tutorial or Something. It will be a Big Help.
Try this:
- (void)webViewDidFinishLoad:(UIWebView *)webView
{
NSString* title = [webView stringByEvaluatingJavaScriptFromString: #"document.title"];
navbar.title = title;
}
Taken from: http://blog.mcohen.me/2008/12/12/getting-the-title-of-a-web-view-in-cocoa-touch/
If you don't want to display the webview, make it hidden and use above delegate method to get what you want...
A way you could do it is by downloading the HTML content at the target url and parse it, for example using hpple.
Example (not tested):
NSURL *url = [NSURL URLWithString:#"http://..."];
//Only using this for simplicity of the example.
//You should really use an asynchronous method instead (see `NSURLSession`).
NSData *urlData = [NSData dataWithContentsOfURL:url];
//Parse HTML and search for title element
TFHpple *hppleParser = [[TFHpple alloc] initWithHTMLData:data];
TFHppleElement *titleElement = [hppleParser search:#"//title"][0];
NSString titleString = titleElement.text;
I'm using the CocoaLibSpotify library to develop an iOS application that will utilize the Spotify API. I've got it just about where I want it, but I've run into a bit of a problem.
When the user touches my "Logout of Spotify" button, I execute the following code:
-(IBAction)logoutButtonTouched:(id)sender
{
// Clear out the user's settings that I am saving.
NSString *appDomain = [[NSBundle mainBundle] bundleIdentifier];
[[NSUserDefaults standardUserDefaults] removePersistentDomainForName:appDomain];
[[SPSession sharedSession] logout:^(void) {
SPLoginViewController *controller = [SPLoginViewController loginControllerForSession:[SPSession sharedSession]];
controller.allowsCancel = NO;
}];
}
This does indeed logout the user and display the SPLoginViewController, but my problem is, the username and password field still contain the values that they'd logged-in with. Does anyone know of a way to clear these fields when I display the SPLoginViewController?
This functionality isn't in the login controller, which is indeed a bug.
You can do it like this. Please note that this is really fragile code and will fail if any internal detail of the login controller changes, and it will in the future:
SPLoginViewController *controller = [SPLoginViewController loginControllerForSession:[SPSession sharedSession]];
id internalLoginViewController = [controller.viewControllers objectAtIndex:0];
UITextField *loginField = [internalLoginViewController valueForKey:#"usernameField"];
UITextField *passwordField = [internalLoginViewController valueForKey:#"passwordField"];
loginField.text = #"";
passwordField.text = #"";
One of the functions of an app I am making involves logging the user into our Campus Portal website. For ease of use, the user may enter a username and password into the app once, and it will be saved for all future log-ins. When the user clicks a button, the information will automatically be sent to log in, and the website will be displayed in a UIWebView.
Let us say that the username and password are each stored in an NSString. How can I use this data to log in to this website programmatically and display the page it in a UIWebView?
I know little about posting and forms, so any help is appreciated.
Would something like this Stackoverflow answer help?
Here's the shell of my code for this
- (IBAction)btnGo:(id)sender {
username = usernameField.text;
password = passwordField.text;
if (saveSwitch.isOn) {
//Save data if the user wants
AppDelegate *appDelegate = (AppDelegate *)[[UIApplication sharedApplication] delegate];
usernameSaved = username;
passwordSaved = password;
[appDelegate.listOfPortalCredentials replaceObjectAtIndex:0 withObject:usernameSaved];
[appDelegate.listOfPortalCredentials replaceObjectAtIndex:1 withObject:passwordSaved];
}
//Insert username and password to web form, log in to portal
//This is where I need help
}
Edit:
Thanks to Karthik I was able to find the HTTP Post using Firebug. It gave me:
appName=ridgefield&portalUrl=portal%2Fridgefield.jsp%3F%26rID%3D0.18423783694092&username=<username>&password=<password>&B1=Log+In&url=portal%2Fmain.xsl%3FrID%3D0.6845596700302482&lang=en
where and represent the real username and password. I believe this is what I need. Using this, how can I display the logged-in page in a UIWebView? Do I just need to load the above URL in the UIWebView?
Use UIWebView, and do a http POST to https://ic.ridgefield.org/campus/verify.jsp with username and password.
To understand how it works, install Firebug on Firefox browser and go to 'Net' tab on firebug, and then open your website and enter some username/password.
You should mimic that action using code. (I always get invalid username or password response from server, coz i dont have an account and i try some random ones, and since there is no signup on the site, there is no way for me to verify this)
UIWebView* webView = [[UIWebView alloc] initWithFrame:self.view.frame];
[self.view addSubview:webView];
NSString* username = #"";
NSString* password = #"";
NSURL* url = [NSURL URLWithString:#"https://ic.ridgefield.org/campus/verify.jsp"];
NSString* body = [NSString stringWithFormat:#"appName=ridgefield&username=%#&password=%#", username, password];
NSMutableURLRequest* request = [NSMutableURLRequest requestWithURL:url cachePolicy:NSURLRequestReloadIgnoringLocalAndRemoteCacheData timeoutInterval:30];
request.HTTPMethod = #"POST";
request.HTTPBody = [body dataUsingEncoding:NSStringEncodingConversionAllowLossy];
[webView loadRequest:request];
So I'm in the progress of developing an app but I'm kinda stuck at the moment. I have 2 textfields right now that the user will have to enter:
Ordernr:xxxxxxxx
Referencenr:xxxxxxx
these 2 values have to be added to this link:
http://www.artis-web.de/cgi-bin/WebObjects/Artis.woa/wa/detailedTrack?referencenr=xxxxxxxx&ordernr=xxxxxxxxxx
now I want safari to open this link. The ideal situation would be the user entering the 2 textfields values and then pressing a button called "open in safari"
What would be the best way to implement this?
Just add the following code to your button action:
NSString *urlString = [NSString stringWithFormat:#"http://www.artis-web.de/cgi-bin/WebObjects/Artis.woa/wa/detailedTrack?referencenr=%#&ordernr=%#", referenceNrTextField.text, orderNrTextField.text];
[[UIApplication sharedApplication] openURL: [NSURL URLWithString: urlString]];
To create the URL you can use...
NSURL *url = [NSURL urlWithString:[NSString stringWithFormat:#"http://www.artis-web.de/cgi-bin/WebObjects/Artis.woa/wa/detailedTrack?referencenr=%#&ordernr=%#", referenceTextField.text, ordernoTextField.text]];
To open in safari...
[[UIApplication sharedApplication] openURL:url];
Check whether the both fields have data and then hit the url with the given data.This condition would be sufficient.
if (Ordernr.text.length!=0 && Referencenr.text.length!=0)
{
NSLog(#"Hit the Url with the entered data");
}
else
{
NSLog (#"Show Error Alert Message");
}
I use the following code in my iOS app to use Instagram iPhone hooks to post a photo to Instagram. I only want the "Open In..." menu to have Instagram app, no other apps. But in my case Camera+ also shows up. How can I restrict to Instagram?
Also, can I directly open Instagram instead of showing Open In menu?
NSURL *instagramURL = [NSURL URLWithString:#"instagram://app"];
if ([[UIApplication sharedApplication] canOpenURL:instagramURL]) {
//imageToUpload is a file path with .ig file extension
self.documentInteractionController = [UIDocumentInteractionController interactionControllerWithURL:[NSURL fileURLWithPath:imageToUpload]];
self.documentInteractionController.UTI = #"com.instagram.photo";
self.documentInteractionController.annotation = [NSDictionary dictionaryWithObject:#"my caption" forKey:#"InstagramCaption"];
[self.documentInteractionController presentOpenInMenuFromBarButtonItem:self.exportBarButtonItem animated:YES];
}
BTW Instagram added an exclusive file extention (ig) and UTI (com.instagram.exclusivegram) for this. It still opens the Open with... menu but the only option is Instagram.
More info here: https://instagram.com/developer/mobile-sharing/iphone-hooks/
You can get the solution from this link.
Save image with the .igo extension instead of .ig. This is the "exclusive" version of the filetype.
Create a UIDocumentInteractionController, then assign the value com.instagram.exclusivegram to the property UTI.
Present your UIDocumentInteractionController with presentOpenInMenuFromRect:inView:animated.
This worked for me, do it like this and you will have only Instagram as the exclusive app to open your image.
NSString *documentDirectory = [NSHomeDirectory() stringByAppendingPathComponent:#"Documents"];
// *.igo is exclusive to instagram
NSString *saveImagePath = [documentDirectory stringByAppendingPathComponent:#"Image.igo"];
NSData *imageData = UIImagePNGRepresentation(filteredImage);
[imageData writeToFile:saveImagePath atomically:YES];
NSURL *imageURL=[NSURL fileURLWithPath:saveImagePath];
_docController=[[UIDocumentInteractionController alloc]init];
_docController.delegate=self;
_docController.UTI=#"com.instagram.photo";
[_docController setURL:imageURL];
_docController.annotation=[NSDictionary dictionaryWithObjectsAndKeys:#"#yourHashTagGoesHere",#"InstagramCaption", nil];
[_docController presentOpenInMenuFromRect:CGRectZero inView:self.view animated:YES];
To answer only your first question: you may probably be able to restrict the "Open in ..." menu to just showing Instagram for your device (by deleting the Camera+ App, for example), but you won't be able to restrict users that install your app to their devices. And that's because the iPhone recognizes which applications are able to open a specific kind of files and it automatically show every one that does.
self.documentInteractionController = [self setupControllerWithURL:imgurl usingDelegate:self];
self.documentInteractionController=[UIDocumentInteractionController interactionControllerWithURL:imgurl];
self.documentInteractionController.UTI = #"com.instagram.exclusivegram";
use this code in same sequence .
here documentInteractionController is object of UIDocumentInteractionController.just for your knowledge.
you will get instagram only in "open in" window.