RegEx validation in XML schema -

- 

  & lt; xs: सरल टाइप नाम = "व्यक्तिगत नाम" & gt; & Lt; xs: प्रतिबंध आधार = "xs: स्ट्रिंग" & gt; & Lt; xs: पैटर्न मान = "। {0,0} | \ w + ([-] \ w +) * ((w) +) +" / & gt; & Lt; / XS: प्रतिबंध & gt; & Lt; / XS: simpleType & gt;   
 मैं अपने XML इनपुट मानों को उपरोक्त पैटर्न में प्रतिबंधित करने की कोशिश कर रहा हूं। यह दो नामों के बीच हाइफ़ेन स्वीकार कर सकता है।  
 यह "अंकित बंसल" जैसे नामों को छोड़ देना चाहिए या पैटर्न में तर्क के अनुसार "अंकित-बंसल अग्रवाल" 0 आर "अंकित बंसल बंसल"  
 लेकिन जब भी मैं निम्नलिखित पैटर्न से चलाता हूं तो मुझे एक त्रुटि देता है जैसे  
 
 [सीडीएटीए ['व्यक्तिगत नाम' तत्व का डेटा प्रकार के अनुसार एक अमान्य मूल्य है। एक त्रुटि हुई, (1, 32 9)।]    
 
 
  ^ ( (?: (?: \ B [a-za-z0- 9] + [-]) +) \ w +) $   
 आप यह कोशिश कर सकते हैं। डेमो देखें।

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

javascript - How to ignore requestAnimationFrame in browsers which don't support it -

-
Assume that we have a feature / module, which increases a website so it is not really necessary and it I use the requestAnimationFrame , which is like IE8 / 9, I could have polyphil requestAnimationFrame , but because it is only an increase, the old browser should ignore it. The code of this module looks more or less like this: ; {Function (window, document, undefined) {'strict use'; // example module // constructor function module, option) {// ...} Module.prototype = {init: function () {// if requestAnimationFrame Not supported, a warning warning ("init") should be popped up;}, method: function () {/ if the request animationfreck is not supported, then alert any warning ("Start");}}; Window.Module = Module;} (window, document) should pop up; Then I can create a new example var example = new module (document.getElementById ('test')); and "interact with" instance.init (); Instance.method (); The problem with this code ...
Read more