Choosing specific array in javascript using value from html select

Question

I have a number of arrays that contain multiple values.

 var ab = [".inb", ".bhm", ".bab", ".mdr"];
 var bc = [".thc", ".lsl", ".cma", ".vth"];
 var de = [".myh", ".rte", ".bab", ".tzs"];
    etc

I am using a select to specify which array to iterate through changing css values on a menu

 $('#browse2').on('change', function () {

    // the value of #browse2 would be ab or bc or de etc
    var fv = $('#browse2').val();

    for (i = 0; i < fv.length; ++i) {
       $(fv[i]).css('opacity', '1.0');
    }

 });

But unfortunately this code only goes through the fv value itself (ie: ab), it doesn't call the array with the same variable name. How do I tell the for statement to use the variable with the same name as the fv value?

Either you will have to use if conditions to decide which array to use, or you can add all arrays in a object and try object[arrayName]. — Rajesh Dixit
– Rajesh Dixit, Commented Nov 4, 2015 at 11:49

Mark Schultheiss · Accepted Answer · 2015-11-04 12:19:20Z

2

I would advice you to store arrays as properties of an object. You will be able to access them with the help of bracket notation:

var obj = {
     ab: [".inb", ".bhm", ".bab", ".mdr"],
     bc: [".thc", ".lsl", ".cma", ".vth"],
     de: [".myh", ".rte", ".bab", ".tzs"]    
 }; 

 $('#browse2').on('change', function () {   
    var fv = $('#browse2').val();

    for (i = 0; i < obj[fv].length; ++i) {
       $(obj[fv][i]).css('opacity', '1.0');
    }

 });

edited Nov 4, 2015 at 12:19

Mark Schultheiss

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answered Nov 4, 2015 at 11:48

Alexandr Lazarev

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2 Comments

MrCycling Over a year ago

Thanks to both Lazarev and Mark. With Mark's correction, this works nicely.

Alexandr Lazarev Over a year ago

Sorry, I've missed it. Glad, to help you!

Curtis · Accepted Answer · 2015-11-04 11:48:53Z

0

Change this line:

var fv = $('#browse2').val();

To:

var fvValue = $('#browse2').val();
var fv;
switch(fvValue){
   case "ab": fv = ab; break;
   case "bc": fv = bc; break;
   case "cd": fv = cd; break;
}

Basically in fv you were storing a string of "ab" rather than a reference to the ab array object.

answered Nov 4, 2015 at 11:48

Curtis

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