What selector can I use to exclude IDs with a suffix?












0














I have a bunch of divs on a page (bg_1, bg_5, bg_112 etc.) that I'd like to assign the style 'display:inline-block'.



However, I have other divs called bg_1_log, bg_5_log, bg_112_log etc..) that I do not want to assign this style to.



I have tried:



 $('[id^=bg_]').attr('style','display:inline-block');


This works, but of course that assigns the style to all divs that start with 'bg_', including bg_112_log...



How can I "assign a style to all divs that are just bg_+number?



I'm using jQuery so either that or JS would be fine.






javascript jquery







share|improve this question




















  • 1




    can you assign a class to them while generating the html?
    – Daniel A. White
    Nov 13 at 18:35
















0














I have a bunch of divs on a page (bg_1, bg_5, bg_112 etc.) that I'd like to assign the style 'display:inline-block'.



However, I have other divs called bg_1_log, bg_5_log, bg_112_log etc..) that I do not want to assign this style to.



I have tried:



 $('[id^=bg_]').attr('style','display:inline-block');


This works, but of course that assigns the style to all divs that start with 'bg_', including bg_112_log...



How can I "assign a style to all divs that are just bg_+number?



I'm using jQuery so either that or JS would be fine.






javascript jquery







share|improve this question




















  • 1




    can you assign a class to them while generating the html?
    – Daniel A. White
    Nov 13 at 18:35














0












0








0







I have a bunch of divs on a page (bg_1, bg_5, bg_112 etc.) that I'd like to assign the style 'display:inline-block'.



However, I have other divs called bg_1_log, bg_5_log, bg_112_log etc..) that I do not want to assign this style to.



I have tried:



 $('[id^=bg_]').attr('style','display:inline-block');


This works, but of course that assigns the style to all divs that start with 'bg_', including bg_112_log...



How can I "assign a style to all divs that are just bg_+number?



I'm using jQuery so either that or JS would be fine.






javascript jquery







share|improve this question















I have a bunch of divs on a page (bg_1, bg_5, bg_112 etc.) that I'd like to assign the style 'display:inline-block'.



However, I have other divs called bg_1_log, bg_5_log, bg_112_log etc..) that I do not want to assign this style to.



I have tried:



 $('[id^=bg_]').attr('style','display:inline-block');


This works, but of course that assigns the style to all divs that start with 'bg_', including bg_112_log...



How can I "assign a style to all divs that are just bg_+number?



I'm using jQuery so either that or JS would be fine.






javascript jquery




javascript jquery






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 13 at 18:41









isherwood

36.6k1081111




36.6k1081111










asked Nov 13 at 18:33









dijon

149617




149617








  • 1




    can you assign a class to them while generating the html?
    – Daniel A. White
    Nov 13 at 18:35














  • 1




    can you assign a class to them while generating the html?
    – Daniel A. White
    Nov 13 at 18:35








1




1




can you assign a class to them while generating the html?
– Daniel A. White
Nov 13 at 18:35




can you assign a class to them while generating the html?
– Daniel A. White
Nov 13 at 18:35












3 Answers
3






active

oldest

votes


















2














You could use the selector using not and attribute contains selector like:



$('[id^=bg_]').not('[id*=_log]')


Or, you can also use regex as @zfrisch suggested but here's an improved version (Selector not only limitted to div):



$('*').filter(function() { return this.id.match(/bg_d$/) })


Or, even like:



$('*').attr('id').match(/bg_d$/)





share|improve this answer























  • Perfect, I didn't know I could use the .not in conjunction with the contains selector.
    – dijon
    Nov 13 at 21:23



















0














You can use JQuery's .filter method






$('div')

    .filter(function() {

        return this.id.match(/bg_d$/);

    })

    .css("backgroundColor", "green");

div:after {

  content: "test";

}

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

<div id="bg_1"></div>

<div id="bg_2"></div>

<div id="bg_3"></div>

<div id="bg_4"></div>

<div id="bg_5"></div>

<div id="bg_6"></div>

<div id="bg_1_log"></div>

<div id="bg_2_log"></div>

<div id="bg_3_log"></div>

<div id="bg_4_log"></div>

<div id="bg_5_log"></div>

<div id="bg_6_log"></div>





Or as another answer pointed out .not






share|improve this answer





























    0














    If you don't want to select the elements whose id ends with _log then you can use following code: 



    $('[id^=bg_]').not('[id$=_log]').attr('style','display:inline-block');





    share|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      You could use the selector using not and attribute contains selector like:



      $('[id^=bg_]').not('[id*=_log]')


      Or, you can also use regex as @zfrisch suggested but here's an improved version (Selector not only limitted to div):



      $('*').filter(function() { return this.id.match(/bg_d$/) })


      Or, even like:



      $('*').attr('id').match(/bg_d$/)





      share|improve this answer























      • Perfect, I didn't know I could use the .not in conjunction with the contains selector.
        – dijon
        Nov 13 at 21:23
















      2














      You could use the selector using not and attribute contains selector like:



      $('[id^=bg_]').not('[id*=_log]')


      Or, you can also use regex as @zfrisch suggested but here's an improved version (Selector not only limitted to div):



      $('*').filter(function() { return this.id.match(/bg_d$/) })


      Or, even like:



      $('*').attr('id').match(/bg_d$/)





      share|improve this answer























      • Perfect, I didn't know I could use the .not in conjunction with the contains selector.
        – dijon
        Nov 13 at 21:23














      2












      2








      2






      You could use the selector using not and attribute contains selector like:



      $('[id^=bg_]').not('[id*=_log]')


      Or, you can also use regex as @zfrisch suggested but here's an improved version (Selector not only limitted to div):



      $('*').filter(function() { return this.id.match(/bg_d$/) })


      Or, even like:



      $('*').attr('id').match(/bg_d$/)





      share|improve this answer














      You could use the selector using not and attribute contains selector like:



      $('[id^=bg_]').not('[id*=_log]')


      Or, you can also use regex as @zfrisch suggested but here's an improved version (Selector not only limitted to div):



      $('*').filter(function() { return this.id.match(/bg_d$/) })


      Or, even like:



      $('*').attr('id').match(/bg_d$/)






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 13 at 18:55

























      answered Nov 13 at 18:36









      Bhojendra Rauniyar

      50.7k2079124




      50.7k2079124












      • Perfect, I didn't know I could use the .not in conjunction with the contains selector.
        – dijon
        Nov 13 at 21:23


















      • Perfect, I didn't know I could use the .not in conjunction with the contains selector.
        – dijon
        Nov 13 at 21:23
















      Perfect, I didn't know I could use the .not in conjunction with the contains selector.
      – dijon
      Nov 13 at 21:23




      Perfect, I didn't know I could use the .not in conjunction with the contains selector.
      – dijon
      Nov 13 at 21:23













      0














      You can use JQuery's .filter method






      $('div')
      
    .filter(function() {
      
        return this.id.match(/bg_d$/);
      
    })
      
    .css("backgroundColor", "green");

      div:after {
      
  content: "test";
      
}

      <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
      
<div id="bg_1"></div>
      
<div id="bg_2"></div>
      
<div id="bg_3"></div>
      
<div id="bg_4"></div>
      
<div id="bg_5"></div>
      
<div id="bg_6"></div>
      
<div id="bg_1_log"></div>
      
<div id="bg_2_log"></div>
      
<div id="bg_3_log"></div>
      
<div id="bg_4_log"></div>
      
<div id="bg_5_log"></div>
      
<div id="bg_6_log"></div>





      Or as another answer pointed out .not






      share|improve this answer


























        0














        You can use JQuery's .filter method






        $('div')
        
    .filter(function() {
        
        return this.id.match(/bg_d$/);
        
    })
        
    .css("backgroundColor", "green");

        div:after {
        
  content: "test";
        
}

        <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
        
<div id="bg_1"></div>
        
<div id="bg_2"></div>
        
<div id="bg_3"></div>
        
<div id="bg_4"></div>
        
<div id="bg_5"></div>
        
<div id="bg_6"></div>
        
<div id="bg_1_log"></div>
        
<div id="bg_2_log"></div>
        
<div id="bg_3_log"></div>
        
<div id="bg_4_log"></div>
        
<div id="bg_5_log"></div>
        
<div id="bg_6_log"></div>





        Or as another answer pointed out .not






        share|improve this answer
























          0












          0








          0






          You can use JQuery's .filter method






          $('div')
          
    .filter(function() {
          
        return this.id.match(/bg_d$/);
          
    })
          
    .css("backgroundColor", "green");

          div:after {
          
  content: "test";
          
}

          <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
          
<div id="bg_1"></div>
          
<div id="bg_2"></div>
          
<div id="bg_3"></div>
          
<div id="bg_4"></div>
          
<div id="bg_5"></div>
          
<div id="bg_6"></div>
          
<div id="bg_1_log"></div>
          
<div id="bg_2_log"></div>
          
<div id="bg_3_log"></div>
          
<div id="bg_4_log"></div>
          
<div id="bg_5_log"></div>
          
<div id="bg_6_log"></div>





          Or as another answer pointed out .not






          share|improve this answer












          You can use JQuery's .filter method






          $('div')
          
    .filter(function() {
          
        return this.id.match(/bg_d$/);
          
    })
          
    .css("backgroundColor", "green");

          div:after {
          
  content: "test";
          
}

          <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
          
<div id="bg_1"></div>
          
<div id="bg_2"></div>
          
<div id="bg_3"></div>
          
<div id="bg_4"></div>
          
<div id="bg_5"></div>
          
<div id="bg_6"></div>
          
<div id="bg_1_log"></div>
          
<div id="bg_2_log"></div>
          
<div id="bg_3_log"></div>
          
<div id="bg_4_log"></div>
          
<div id="bg_5_log"></div>
          
<div id="bg_6_log"></div>





          Or as another answer pointed out .not






          $('div')
          
    .filter(function() {
          
        return this.id.match(/bg_d$/);
          
    })
          
    .css("backgroundColor", "green");

          div:after {
          
  content: "test";
          
}

          <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
          
<div id="bg_1"></div>
          
<div id="bg_2"></div>
          
<div id="bg_3"></div>
          
<div id="bg_4"></div>
          
<div id="bg_5"></div>
          
<div id="bg_6"></div>
          
<div id="bg_1_log"></div>
          
<div id="bg_2_log"></div>
          
<div id="bg_3_log"></div>
          
<div id="bg_4_log"></div>
          
<div id="bg_5_log"></div>
          
<div id="bg_6_log"></div>





          $('div')
          
    .filter(function() {
          
        return this.id.match(/bg_d$/);
          
    })
          
    .css("backgroundColor", "green");

          div:after {
          
  content: "test";
          
}

          <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
          
<div id="bg_1"></div>
          
<div id="bg_2"></div>
          
<div id="bg_3"></div>
          
<div id="bg_4"></div>
          
<div id="bg_5"></div>
          
<div id="bg_6"></div>
          
<div id="bg_1_log"></div>
          
<div id="bg_2_log"></div>
          
<div id="bg_3_log"></div>
          
<div id="bg_4_log"></div>
          
<div id="bg_5_log"></div>
          
<div id="bg_6_log"></div>






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 13 at 18:42









          zfrisch

          4,42311024




          4,42311024























              0














              If you don't want to select the elements whose id ends with _log then you can use following code: 



              $('[id^=bg_]').not('[id$=_log]').attr('style','display:inline-block');





              share|improve this answer


























                0














                If you don't want to select the elements whose id ends with _log then you can use following code: 



                $('[id^=bg_]').not('[id$=_log]').attr('style','display:inline-block');





                share|improve this answer
























                  0












                  0








                  0






                  If you don't want to select the elements whose id ends with _log then you can use following code: 



                  $('[id^=bg_]').not('[id$=_log]').attr('style','display:inline-block');





                  share|improve this answer












                  If you don't want to select the elements whose id ends with _log then you can use following code: 



                  $('[id^=bg_]').not('[id$=_log]').attr('style','display:inline-block');






                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 13 at 18:44









                  Gowri Sankar K

                  1




                  1






























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