I know that I am using a class to add an HTML link, which single Functional at one time: @ html.ActionLink ("add", "updatenot", "notes", new {id = 0, type = (int ), The "{@ class =" BTN BTN-primary icon-edit "}, empty)
However,
is there a way to define a class (Add a style to all divs in a very CSS file like:
div {color: red;}; ^ ^ | | This will work on * all * divs < I can just write
/ Pre>
There will be color: red included
automatically without having to go to each action link and @ class = "myClass" / Code>
for example
define the class for the nonlinear link One way to do this: To add style for all button instances:
input [type = "button"] {background color: red; }
What I could do with something like:
input [type = "actionlink"] {// Akshnlink all projects Style}} and therefore all can be written as aplication:
@html. Exynalink ("Action", "Controller")
And automatically include the style given in my CSS file?
I do not do it the first way to do, but I've already defined as ~ 100 made, no class, and copy and paste:
class = "myClass"
Akshnlink just generate common anchor, then You can type:
a {color: red; } But just to get ActionLinks you will need to call them through a square or container.
Also take a better way to custom Aekshnlink inside a default class to it and using this class you can make your selection, as you will use this new custom Akshnlink and paste Class No need to copy
public static class Linkakstenshn {public static MvcHtmlString MyActionLink (this HtmlHelper html helper, string link text string action, string controller) {var currentAction = htmlHelper.ViewContext.RouteData.GetRequiredString ( "Action" ); Var current controller = htmlHelper.ViewContext.RouteData.GetRequiredString ("controller"); If (Action == Current Actions & Controllers == Current Controller) {var anchor = new TagBuilder ("a"); Anchor. Attribute ["href"] = "#"; Anchor.AddCssClass ("currentPageCSS"); Anchor.SetInnerText (LinkText); Return MvcHtmlString.Create (anchor.ToString ()); } Return htmlHelper.ActionLink (linkText, Action, Controller); }}% = HTML.MyActionLink ("Hello Foo", "Index", "Home")%> & Lt;% = HTML.MyActionLink ("Hello Bar", "About", "Home")%> code has been copied
Popular posts from this blog
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...
सी / सी ++ में एक सरणी के लिए एक विशिष्ट घोषणा है: प्रकार का नाम [तत्व]; जहां एक प्रकार वैध है (जैसे कि इंट, फ्लोट ...), नाम एक मान्य पहचानकर्ता और तत्व फ़ील्ड है (जो हमेशा चौकोर ब्रैकेट्स [] में संलग्न होता है), सरणी की लंबाई को निर्दिष्ट करता है तत्वों की संख्या के संदर्भ में। इसलिए मैं एक सरणी की घोषणा करता हूँ जिसमें 2 तत्व हैं int a [2]; एक [3] = 4; क्यों यह अपवाद नहीं फेंकते? सीमा से बाहर की जाँच कुछ ऐसा है जो आप संभवतः जावा जैसे कुछ उच्च स्तर की भाषा से उपयोग किया जाता है। हालांकि सी / सी ++ में यह डिफ़ॉल्ट रूप से नहीं किया जाता है। यह आपको सरणी की सीमाओं की जांच करने के लिए एक छोटा प्रदर्शन हिट देता है और इसलिए सी तर्क यह है कि आप इसे स्वयं मैन्युअल रूप से कर सकते हैं, जब तक आप इसे सर्वश्रेष्ठ संभव प्रदर्शन की पेशकश कर सकते हैं। सी + + एसटीएल कंटेनर जैसे वेक्टर आमतौर पर एक पर () ऑपरेशन को बाध्य जांच करने के लिए समर्थन देता है और चूंकि आप [] अधिभार लोड कर सकते हैं - आप सरणी के लिए बाउंड-चेक सक्षम कर सकते हैं -स्टाइल एक्सेस। यदि सरणी एक कच्चे पॉइंटर ह...
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
Comments
Post a Comment