Why is the javascript variable undefined? (2nd time around...)

Question

I am reposting this question with revised code that use more of jQuery.

Here is the HTML code that defines the objects:

<LEGEND><b>Select Study Sample</b></LEGEND>
<p>
<P CLASS=select_header>Study - Box - Well - Lab ID<br>
<SELECT multiple size=20 NAME="stylabid" ID="stylabid" onchange=show_stylabid() onclick=clear_fetch() STYLE="width: 115px">
<?php 
  $i=0;
  while ($i < $numstylabids) {
    $styarr = pg_fetch_row($stylabidresult);
    echo "<option value=$styarr[0]>$styarr[1]\n";
    $i++;           
  }
  ?>
</select>

and

<LEGEND><b>Select Project Sample</b></LEGEND>
<p>
<P CLASS=select_header>Project - Box - Well - Lab ID<br>
<SELECT multiple size=20 NAME="pjtlabid" ID="pjtlabid" onchange=show_pjtlabid() onclick=clear_fetch() STYLE="width: 115px">
<?php 
  $j=0;
  while ($j < $numpjtlabids) {
    $pjtarr = pg_fetch_row($pjtlabidresult);
    echo "<option value=$pjtarr[0]>$pjtarr[1]\n";
    $j++;           
  }
  ?>
</select>

Now, here is the javascript; I am simply trying to get the values of the selected object, which can be either a Study or a Project. I use this later to retrieve data from the database:

function fetchgenotype() {
    var ms, mndx, ss, sndex = 0;

    ms = $("#marker")[0].selectedIndex;    

    if (Study_selected) {
      ss = $("#stylabid")[0].selectedIndex;
      sndx = $("#stylabid").val(ss);
    } else
      ss = $("#pjtlabid")[0].selectedIndex;
      sndx = $("#pjtlabid").val(ss);
    }    

    // for debugging...    
 alert('ss='+ss+', sndx='+sndx);

This code dies on the line sndx = ... Would really appreciate any help!

TIA

sndx was never defined. Pretty much what undefined means. From the looks of this, you know better! :) (p.s. - define sndx or use sndex) — Travis J
– Travis J, Commented Sep 21, 2012 at 21:19
When is that function being called? I don't see any references to it in the rest of what you've posted. — Anthony Grist
– Anthony Grist, Commented Sep 21, 2012 at 21:29
What exactly do you mean by This code dies on the line sndx = ... That line shouldn't be causing any errors, even if it doesn't actually return what you expect it to. — user400654
– user400654, Commented Sep 21, 2012 at 21:40
What exactly are you trying to assign sndx? You know .val(value) is a setter method right? — wirey00
– wirey00, Commented Sep 21, 2012 at 21:41

Dr.Molle · Accepted Answer · 2012-09-21 21:30:01Z

1

Currently i don't know what $("#marker") an Study_selected are.

Assuming there is anything OK so far, change the function into:

if (Study_selected) {
      sndx = $("#stylabid").val();
    } else
      sndx = $("#pjtlabid").val();
    }

val() returns the value of a form-element when used without an argument, otherwise it will set the value and return an jQuery-object.

There is no need to retrieve the selectedIndex, you may access the value of the select-element directly(it will be the value of the currently selected option)

answered Sep 21, 2012 at 21:30

Dr.Molle

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4 Comments

rixter Over a year ago

Thanks Dr Molle; that explains a lot. No, I did not know the diff between setter and getter jQ methods; still learning!

rixter Over a year ago

I modified my code per your suggestion: if (Study_selected) { //ss = $("#stylabid")[0].selectedIndex; sndx = $('#stylabid').val(); } else //ss = $("#pjtlabid")[0].selectedIndex; sndx = $('#pjtlabid').val(); } ........ but this line throws an error and now Firebug tells me 'TypeError: $ is not a function' [what the heck does that mean? This is standard jQ notation!]

Dr.Molle Over a year ago

When $ is not a function it may be that you didn't include jQuery.

rixter Over a year ago

I appreciate all the help I got here, but finally saw the error: the if else statement was missing a bracket between else and sndx...duh! Never underestimate the potential to overlook stupid mistakes....;>

wirey00 · Accepted Answer · 2012-09-21 21:42:01Z

1

I think you have a syntax error

sndx should be sndex .. or the other way around

var ms, mndx, ss, sndex = 0; // <--

$("#pjtlabid").val(ss) // <-- setter method

.val(value) is also a setter method so sndx is undefined because it never was defined with any value

edited Sep 21, 2012 at 21:42

answered Sep 21, 2012 at 21:17

wirey00

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Comments

Adriano Carneiro · Accepted Answer · 2012-09-21 21:17:31Z

0

It's a typo:

var ms, mndx, ss, sndex = 0;

should be

var ms, mndx, ss, sndx = 0;

answered Sep 21, 2012 at 21:17

Adriano Carneiro

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5 Comments

Anthony Grist Over a year ago

Calling sndx = ... will happily declare a global variable, even if sndx is undefined prior to that. The typo may cause problems, but it isn't the cause of this problem, assuming all of the information in the question is correct.

user400654 Over a year ago

@AnthonyGrist no... it will be local to the scope due to the var keyword.

Anthony Grist Over a year ago

@KevinB There isn't a var keyword in the code in the question.

user400654 Over a year ago

@AnthonyGrist It's right here, var ms, mndx, ss, sndex = 0; copied from the OP. It's also in Adrian's answer.

Anthony Grist Over a year ago

@KevinB Right, sndex (with an e), and we're talking about a declaration for sndx (no e).

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