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Home/ Questions/Q 7056951
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T03:55:41+00:00 2026-05-28T03:55:41+00:00

Here is my code – fairly simple: $(document).ready(function(){ $(.location).each(function(){ $(this).click(function(){ $(map).hide(); var class =

  • 0

Here is my code – fairly simple:

$(document).ready(function(){         
  $(".location").each(function(){             
    $(this).click(function(){                 
      $("map").hide();                 
      var class = $(this).attr("data");                       
      $(".location").removeClass("activeLocation");                 
      $(this).addClass("activeLocation");                 
      $("." + class).show();             
    })         
  })     
});

HTML

<div id="locations">     
  <div class="location activeLocation" data="brian">1205 S 75th St</div>     
  <div class="location one" data="mark">6603 A Royal Street</div>     
  <div class="location" data="dan">4725 Merle Hay Rd</div>     
  <div class="location" data="andy">62 Soccer Park Road</div>     
  <div id="mapviewer" class="brian map">Map to 1205 S 75th Street</div>     
  <div style="display: none;" id="mapviewer" class="mark map">Map to 6603 A Royal Street</div>     
  <div style="display: none;" id="mapviewer" class="dan map">Map to 4725 Merle Hay Rd</div>    
  <div style="display: none;" id="mapviewer" class="andy map">Map to  62 Soccer Park Road</div>
</div>

Everything works if I comment out two lines:

var class = $(this).attr("data");
$("." + class).show();

Without these two lines commented, no action at all. These two lines of code look OK to me.
What am I missing?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T03:55:42+00:00Added an answer on May 28, 2026 at 3:55 am

    Right after the line

    var class = $(this).attr("data");
    

    do

    console.log(class) 
    

    to see what it contains. Additionally you do not have to wrap this all in an each() function, just change the first line to

    $('.location').click(function() {...
    

    all of the this’s within the function will be this .location object

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