qt quick - I have this version of Qt. still unable to use Qt.controls and QtQuick 2.0. -


मैंने अपने क्यूटी संस्करण को

  • क्यूटी निर्माता 2.7.1
  • क्यूटी 5.0.2 (32 बिट) के आधार पर
  • 6 सितंबर 2013 को 08:23:22 पर निर्मित

    फिर भी मैं क्यूटी प्रयोग नहीं कर पा रहा हूं त्वरित नियंत्रण और क्यूटी क्विक 2.0 मुझे यह त्रुटि मिलती है 

      मॉड्यूल "क्यूटक्वाइक" संस्करण 2.0 आयात स्थापित नहीं है QtQuick 2.0   
     मैंने क्यूटीकिक्क 2.0 आयात किया है, अभी भी यह काम नहीं कर रहा है।   
     
     
     समाधान जोड़ रहा हो सकता है:  
      QT + = quick   
     अपने .pro फ़ाइल में यदि यह मदद नहीं करता है तो कृपया इंस्टॉल करें।

Comments

Post a Comment

Popular posts from this blog

sonata admin - how to add ManyToOne property field in configureListfield in sonataAdmin -

I have a phone unit unit that contains many properties for many unit entity classes I An exception is given during the rendering of a template ("You must define a associated_property option or Create a \ __counter __ \ Examens \ ExamensBundle \ unit \ Grade Proxy :: __ toString service examens.examens.admin.fonctionnaire from the field option idGrade to the method is in SonataDoctrineORMAdminBundle "): CRUD: Line_ on line_list_orm_many_to_one.html.tw Fontionnaire.php class: & lt ;? Php namespace exams \ ExamensBundle \ Entity; Use the theory / ORM \ mapping as an ORM; / ** * PhoneNationLayer * / Class Fictionalair {//...//// ** * @VARAEfimens \ Examinations Bundle \ Unit \ Grade * / Private $ IDGed; //...//} FonctionnaireAdmin.php: & lt ;? Php namespace exams ExamensBundle \ Admin; Use Sonata \ AdminBundle \ Admin \ Admin; Sonata \ AdminBundle \ Datagrid \ ListMapper; Sonata \ AdminBundle \ Datagrid \ DatagridMapper; Sonata \ AdminBundle \ Validat...
Read more

wpf - ControlTemplate with changes for every class -

I have the following ControlTemplate that is the same for all my custom controls. & amp; the lift; ControlTemplate x: key = "PssFunctionControlBaseHorizontal" TargetType = "Local: PssFunctionControlBase" & gt; & Lt; Grid & gt; & Lt; Threshold Threshold = "1" & gt; & Lt; Content Control X: Name = "Inner Content" Template = "{Static Resources Inner Contentbase Horizontal}" /> & Lt; / Border & gt; & Lt; Control x: name = "PART_ResizeDecorator" visibility = "short" template = "{static resource resize deccurratematet}" /> & Lt; / Grid & gt; & Lt; ControlTemplate.Triggers & gt; & Lt; Data Trigger Value = "True" Binding = "{Binding Editing Moded, Relative Soros = {ResolveSource Search Engineer, Instant Type = {x: Type Local: PssViewLayoutControl}}}" & gt; & Lt; Setter target name = "PART_ResizeDecorator" ...
Read more

python - how we can use ajax() in views.py in django? -

I have a little code here. Could you please tell me what this code really does? please. Here in the AJAX call: $ .ajax ({url: "{% url} upload_document '%}", type: "POST", data : {Csrfmiddletoken: document.getElementsByName ('csrfmiddletoken') [0] Price, Title: document.getElementById ('title'). Value, // document: document: document.getElementById ('document'),}, datatype: "Jason", success: function (feedback) {if (feedback == "true") {// success} other {// attachment errors}}}); How AJAX works in the Django and how we can see the AJAX request. Here is the idea of ​​my thoughts def upload_document (requested): Print request. POTT print request FILES if request.is_ajax (): If request.method == 'POST': form = UploadForm (request.POST, request.FILES, user = request.user) if form.is_valid (): Form.save () return HTTPPCS (SimpleJason Dumps ('True'), Mime Type = 'App / Jason') Other:...
Read more