How can I check text content with respect to display:none style

Question

I would like to get text content of a DOM node, but without elements that are effectively hidden (have style display:none ).

I know about DOM textContent property and jQuery's text() method (I think it uses textContent anyway), but it doesn't respect display:none :

var div = $("<div><span style='display:none'>not </span>good</div>");
var text = div.text(); // "not good"
var textContent = div[0].textContent; // "not good"

How can I get just "good" content from that div?

I need this unit test my AngularJS directive. I am writing tests using Jasmine and my test is something similar to that:

    it('should check text ignoring hidden content', function() {
      $rootScope.hidden = true;
      var template = '<div><span ng-hide="hidden">not </span>good<div>';
      var element = $compile(template)($rootScope);
      $rootScope.$digest();

      console.debug(element.html());    // prints: '<span class="ng-hide" ng-hide="hidden">not </span>good<div></div>'
      console.debug(element.text());    // prints: 'not good'
      expect(element.text()).toEqual('good');   // oops, that fails :(
});

In my real test I want to test my custom directive instead of ng-hide , but the actual directive is irrelvant for that question.

I am completely new to Jasmine and Javascript unit testing, so please also let me know if this is just not the way to go.

Answer 1

A few considerations to keep in mind: you can use clone() to avoid modifying the original object, however to use :visible , the elements have to actually be in the DOM.

 var sample = $("<div><span style='display:none'>not </span>good <div>content</div></div>"); var t = sample.clone(); $('body').append(t); t.find('*:not(:visible)').remove(); t.remove(); alert(t.text()); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 

Answer 2

if you apply a class for your style like .noDisplay { display:none; } you could do something like div.find('span.noDisplay').remove(); with jQuery

Answer 3

Try The following. It will get the text of only "Visible" spans.

 var text=""; $("#content-div").find('span:visible').each(function(){ text=text+' '+$(this).text(); }); alert(text); 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="content-div"><span >ok</span> <span style="display:none">hidden</span></div> 

Answer 4

尝试这个

 $('div').find('span').text("").parent().text();

Answer 5

You can use jquery remove() function and apply to the div children element.

Add div.children().remove(); right after var div = $("<div><span style='display:none'>not </span>good</div>"); and you will get only "good" content

Answer 6

I guess you need to filter the contents and you should target the nodeType = 3 which is assigned for TEXT_NODE .

 var div = $("<div><span style='display:none'>not </span>good</div>"); var textContent = div.contents().filter(function() { return this.nodeType == 3; }).get(0).textContent; // "good" $('body').html(textContent); // logs "good" 

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>