ios - Displaying the Initial View Controller as a Form Sheet -

-

What is the correct way to display the initial view controller as a form sheet? Background: My initial screen is a login controller that after logging in to the map when the user logs out, it presents the login screen as an entry sheet, but when the application launches It's in full screen.

UPDATE: I used the following code to make it work: UIViewController * loginViewController = [self.storyboard instantiateViewControllerWithIdentifier: @ "LoginController"]; LoginViewController.modalPresentationStyle = UIModal PresentationFormatate; [Self-current ViewController: Animated login viewer controller: not complete: zero];

Show an empty view controller and display your login controller as a form sheet . 

  // In your initial view controller - (zero) viewDidload {loginviewcroller * log inview watcher controller = // you can login login login login viewcontroller.modalPresentationStyle = UIModalPresentationFormSheet; [Self-Current ViewController: Animated ViewController: Not Complete: Zero]}

Comments

Post a Comment

Popular posts from this blog

java - ImportError: No module named py4j.java_gateway -

-
I'm trying to call the Python using the Java program py4j . I've been installing the plug-in Eclipse and test name Piidvi project. I'm trying to execute the following part of the code found on py4j webpage: Import from py4j.java_gateway to JavaGateway, java_import gateway = JavaGateway () jvm = gateway.jvm java_import (jvm, '' Org.eclipse Kkorkrisorsej. * ") Vrkspes_rut = Jvankresourkesplginkgetvrkspas (). GetRoot (Gateway .help (workspace_root, '* Projects *') project_names = [project.getName () (for projects workpace_root.getProjects))] print (Projekt_nam) But I There is an error in import. I have checked that the P4JJ is present in the Jar Eclipse plugin directory. Can anyone help please? I had to install the py4j application
Read more

python - Receiving "KeyError" after decoding json result from url -

-
I am new to Python I am trying to parse JSON result from a URL. Basically, I was using the following: response = urllib.request.urlopen (url) json_obj = json.load (response) It should be a stroke "str 'not' bytes' in the lines of a given" JSON object, so after searching on the StackoverView Flo, I decode the response in this way: F = urllib.request.urlopen (Url) charset = f.info (). Get_param ('charset', 'utf8') data = f.read () decoded = json.loads (data.decode (charset)) If I print "decode" I is as follows: { 'link': { 'summary data': 'https: // localhost / piwebapi / streams / p0_7qHaW4UHU-RlCaz8tpasAAQAAAAU0hJTExNQU42NDIwXFNJTlVTT0lE / summary' 'value': 'https: // localhost / Piwebapi / streams / P0_7qHaW4UHU-RlCaz8tpasAAQAAAAU0hJTExNQU42NDIwXFNJTlVTT0lE / price ',' InterpolatedData ':' https: // localhost / Piwebapi / streams / P0_7qHaW4UHU-RlCaz8tpasAAQAAAAU0hJTE...
Read more

python - 'NoneType' object is not callable using 'find_all' in BeautifulSoup -

-
Here's my code using find_all, but this is my code. Works great with: (): To get beautiful suits R = requests from beautiful soup imports to import. (URL_DEFINED) Print Rustusodod Soup = Beautiful Soup (Rtext) Print soup.fund_all ('ul') What I found: Tracebacks (Most recent call end): File "scraper.py", line 1 9, & lt; Module & gt; Print soup.find_all ('ul') TypeError: The 'NoneType' object is not worth the corner It seems that you are using Sundersup version 3, in which a slightly different naming convention is used, such as: .findAll , while SunderSupo 4 standardized naming, like more PEP8, Such as: .find_all (but retains the old naming for previous compatibility) Note that soup ('ul') is equal to both search for all . To download and install, install pipes PrettySpeed ​​4 . Then to change your import: import from bs4 BeautifulSoup Then you are good go.
Read more