javascript function , if input change

Question

I coded this today, but when i use my dropdown i doesn't work... it works fine when I'm writing the numbers myself. Why?!

  function favBrowser() {
    var mylist=document.getElementById('myList');
    document.getElementById('favorite').value=mylist.value;
}
 function bar() {
document.write('Hello World!');


<select id="myList" onchange="favBrowser()">
    <option></option>
    <option value="a">a</option>
    <option value="b">b</option>  
    <option value="c">c</option>
    <option value="d">d</option>
    <option value="e">e</option>
</select>
<p>heheheh: <input type="text" id="favorite" size="20" oninput="return bar()" onValueChanged="return bar()" value=""></p>

Don't use document.write after the document has been loaded/rendered. developer.mozilla.org/en-US/docs/DOM/document.write — Ian
– Ian, Commented Mar 20, 2013 at 14:38

believe me · Accepted Answer · 2013-03-20 14:43:07Z

1

I think this is what you're looking for

function favBrowser(list) {
    document.getElementById('favorite').value=list.value;
}


<select id="myList" onchange="favBrowser(this)">
    <option></option>
    <option value="a">a</option>
    <option value="b">b</option>  
    <option value="c">c</option>
    <option value="d">d</option>
    <option value="e">e</option>
</select>
<p>heheheh: <input type="text" id="favorite" size="20" value=""></p>

answered Mar 20, 2013 at 14:43

believe me

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2 Comments

user2062819 Over a year ago

when i use that my dropdown doesnt print de value from dropdown , only undefined ? ;o

believe me Over a year ago

You mean the input value doesn't print the value from the dropdown? Can you please show the onchange you used?

Laurent S. · Accepted Answer · 2013-03-20 14:40:42Z

0

select elements don't have a value, option elements in it have a value. Try the following instead :

document.getElementById('favorite').value=mylist.options[myList.selectedIndex].value;

not tested, but that's the idea

answered Mar 20, 2013 at 14:40

Laurent S.

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Comments

Ihor Patychenko · Accepted Answer · 2013-03-20 14:49:52Z

0

Having this code html code:

<select id="myList" onchange="favBrowser(this.value)">
    <option></option>
    <option value="a">a</option>
    <option value="b">b</option>  
    <option value="c">c</option>
    <option value="d">d</option>
    <option value="e">e</option>
</select>

<input type="text" id="favorite" size="20" value=""/>

You can do the next:

function favBrowser(val) {
     document.getElementById('favorite').value=val;
}

Every time that you changes the value of drop down list, automaticly changes the value of input because, you are sending the value of drop down list as param.

answered Mar 20, 2013 at 14:49

Ihor Patychenko

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4 Comments

user2062819 Over a year ago

my problem is , when my dropdown changes the value nothing happens , but when i write my own value in de input document.write('Hello World!'); displays

Ihor Patychenko Over a year ago

Dont use document.write after the page loaded.

user2062819 Over a year ago

document.write was only a thing i wrote to se the change when input has an value :)

Ihor Patychenko Over a year ago

but document.write overwrite all data that you had in a document

Reinstate Monica -- notmaynard · Accepted Answer · 2013-03-20 15:00:10Z

0

First, there is no event onValueChanged; you're looking for onchange.

The reason why bar() is not being triggered when your drop-down script changes the value of favorite is because oninput is triggered when the value of the element is changed through the user interface (not a script), and onchange is triggered when blur occurs (that is, the input element loses focus). So, in order to get the results you want, you'll need to call bar() from favBrowser():

function favBrowser() {
  var mylist=document.getElementById('myList');
  document.getElementById('favorite').value=mylist.value;
  bar();
}

answered Mar 20, 2013 at 15:00

Reinstate Monica -- notmaynard

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2 Comments

user2062819 Over a year ago

hmmm , can u write an example? I'm new on this :(

Reinstate Monica -- notmaynard Over a year ago

The point is, you can't use oninput or onchange to do something after a script changes the value of an element like you're trying to do. Instead you need to just have the script do it.

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