Javascript undefined function.. but it really is defined

Question

I have a few functions running on my page. The first one is called by an onclick event from a select field this function then calls another function zip() this function runs a small script. If the user leaves a field blank then the function should be recalled by the checkit() function as a validation method it is returning undefined on the second call.. The first call from colorchange() works..

function colorchange(value){
    if (value != ''){
        document.getElementById('delivery').style.borderColor="#000000";
    }
    else{
        document.getElementById('delivery').style.borderColor="#ff0000";
    }
    if (value == 'deliver'){
    zip();
}
}

function checkit(){

    var x=document.forms["pricing_form"]["delivery"].value;
        if (x==null || x=="")
          {
          alert("Please Choose Delivery Method");
          return false;
          }

        if (document.getElementById('delivery').value == 'deliver'){
            alert ('deliver');
            zipcode = document.getElementById('zip') == '';

            if (zipcode == '' || zipcopde == null || zipcode == 'false'){
                alert('please enter a valid zip code');
                zip();
                return false;
            }
        }
}

function zip(){

    var zip = prompt("Ship-To Zip Code", "");

    if (zip != null && zip != ''){
        document.getElementById('zip').value=zip;
        }
        alert (document.getElementById('zip').value);

}

I renamed the fields as follows to get this script to function properly

function colorchange(value){
    if (value != ''){
        document.getElementById('delivery').style.borderColor="#000000";
    }
    else{
        document.getElementById('delivery').style.borderColor="#ff0000";
    }
    if (value == 'deliver'){
    zip_get();
}
}

function checkit(){

    var x=document.forms["pricing_form"]["delivery"].value;
        if (x==null || x=="")
          {
          alert("Please Choose Delivery Method");
          return false;
          }

        if (document.getElementById('delivery').value == 'deliver'){
            alert ('deliver');
            zipcode = document.getElementById('zip') == '';

            if (zipcode == '' || zipcode == null || zipcode == 'false'){
                alert('please enter a valid zip code');
                zip_get();
                return false;
            }
        }
}

function zip_get(){

    var zip_entry = prompt("Ship-To Zip Code", "");

    if (zip_entry != null && zip != ''){
        document.getElementById('zip').value=zip;
        }
        alert (document.getElementById('zip').value);

}

Typo found: zipcoPde at zipcode == '' || zipcopde == null || zipcode == 'false' — biziclop
– biziclop, Commented Dec 8, 2011 at 18:01

Dave Newton · Accepted Answer · 2011-12-08 17:59:40Z

3

if (zipcode == '' || zipcopde == null || zipcode == 'false'){

You sure it's the method that's undefined, or zipcopde?

I'm also pretty sketched out by naming functions the same as variables--much more dangerous in JavaScript than in some other languages; maybe don't. Should be okay here, but still.

answered Dec 8, 2011 at 17:59

Dave Newton

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2 Comments

Daniel Hunter Over a year ago

your right.. Ill change the variable and see if that helps.. also, i put an alert in the if statement and it will run, developer tools says zip() is undefined, and the return false is ignored

Dave Newton Over a year ago

@DanielHunter If they're all on the same page, and you don't re-assign zip somewhere else (you're missing at least one var in checkit, for example), not sure how it could break.

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