CODEWITHSUNDEEP
javascripthtmlforms
de1337ed
de1337ed

Reputation: 3315

Autofill form completion in javascript

I have a checkbox that calls a javascript function. When this checkbox is checked, other values in a form are auto filled out. The type of these other values are another checkbox and an option select field. I have the following in html:

<div id="NA" >
<input type="checkbox" name="remove_cd" value="r_on" id="r_on_cd" />N/A:
<select name="reason" id="reason_list">
<option value="1">Option 1</option>
<option value="2">Option 2</option>
<option value="3">Option 3</option>
</select>
<font color="#cc0000">Reason Required</font><hr/></div>

In javascript, I have the following in a function:

...
var y = document.getElementById("NA").children;
for(var i=0; i<y.length; y++){
    y[i].checked=true;
    y[i].options.selectedIndex=2;
}
...

I'm a little confused as to why this isn't working. When I click on a checkbox in the form, the checkbox under the <div id="NA"> gets checked, but the option in the dropdown doesn't get changed. Ideas?

Upvotes: 1

Views: 912

Answers (3)

Krasimir
Krasimir

Reputation: 13529

I'm not sure what exactly you want to do, but try this:

<div id="NA" >
<input type="checkbox" name="remove_cd" value="r_on" id="r_on_cd" />N/A:
<select name="reason" id="reason_list">
<option value="1">Option 1</option>
<option value="2">Option 2</option>
<option value="3">Option 3</option>
</select>
<font color="#cc0000">Reason Required</font><hr/></div>
<script>
    window.onload = function() {
        var y = document.getElementById("NA").children; // it's not a good idea to use children array, because if you add a new element in the begining your script will fail
        var checkbox = y[0]; // it is better here to use: document.getElementById('r_on_cd');
        var select = y[1]; // it is better here to use: document.getElementById('reason_list');
        checkbox.onclick = function() {
            select.options.selectedIndex = 2;
        };
    }
</script>

Upvotes: 0

codingbiz
codingbiz

Reputation: 26386

You have y++ instead of i++ in your loop. Try this

for(var i=0; i<y.length; i++){
    y[i].checked=true;
    y[i].selectedIndex=2;

}

calling .selectedIndex on options throws Cannot set property 'selectedIndex' of undefined

Upvotes: 2

Scott Sauyet
Scott Sauyet

Reputation: 50787

y[i].options.selectedIndex=2;

should probably just read:

y[i].selectedIndex=2;

Upvotes: 0

Related Questions