Joining arrays

**bcmann** · 05-14-2009 02:53 PM

Can anyone tell me why this doesn't work?

Code:

        var obj = document.getElementById('form_id');
	var myArray = new Array();
	var _input = obj.getElementsByTagName('input');
	var _select = obj.getElementsByTagName('select');

        myArray = _input.concat(_select);

I'm trying to get an array that contains all the input & select tags. When I run this, myArray contains the input tags, but not the select tags.

Any suggestions would be greatly appreciated.
Brian

**JMRKER** · 05-14-2009 03:11 PM

Only a guess.

It maybe that what is being returned is not an array, but is instead a list of object or element references.

Try putting an alert after collecting the information to see what you have got!
ie:
alert(_input.join('\n'));
alert(_select.join('\n'));

**bcmann** · 05-14-2009 03:17 PM

Thanks for the quick reply!

I tried putting an alert like you suggested, and I'm getting an error that _input.join is not a function. Which tells me that _input is not an array. But when I do "alert(_input.length)" I get an alert that says "2". So is a list of objects not an array? I didn't realize that a list is different that an array. Is there a way to join two lists together?

Thanks again!
Brian

**JMRKER** · 05-14-2009 03:29 PM

How about trying:

Code:

var myArray = new Array();
for (i in _input) { myArray[myArray.length] = i; }
for (i in _select) { myArray[myArray.length] = i; }
alert(myArray.join('\n'));

What happens then?

**JMRKER** · 05-14-2009 09:30 PM

I tried this as a test ... Seems to work without any errors
and it displays something ... I just have no idea what it means.

Code:

<html>
<head>
<title>Concatonate Tags into Array</title>
<script type="text/javascript">
function DisplayTags() {
  var obj = document.getElementById('form_id');
  var _input = obj.getElementsByTagName('input');
  var _select = obj.getElementsByTagName('select');

  var myArray = new Array();
  for (i in _input) { myArray[myArray.length] = i; }
  for (i in _select) { myArray[myArray.length] = i; }
  alert(myArray.join('\n'));
}
</script>
</head>
<body>
<form id="form_id" name="form_id" onsubmit="return false">
 <table border="1"><tr><td>
 <input type="input" name="inA1" value="A"> A<br>
 <input type="input" name="inB1" value="B"> B<br>
 <input type="input" name="inC1" value="C"> C<br>
</td><td> 
 <select id="SBox1" name="SBox1">
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
 </select>
</td><td> 
 <select id="SBox2" name="SBox2">
  <option value="1">1</option>
  <option value="2">2</option>
  <option value="3">3</option>
  <option value="4">4</option>
  <option value="5">5</option>
 </select>
</td><td>
 <input type="input" name="inA2" value="A"> A<br>
 <input type="input" name="inB2" value="B"> B<br>
 <input type="input" name="inC2" value="C"> C<br>
</td></tr></table>
 
 <button onclick="DisplayTags()">Display Tag Array</button>
</form>
</body>
</html>

**bcmann** · 05-15-2009 10:15 AM

LOL.. I tried your code, and I can't figure out what it's outputting either.

I tried substituting "alert(myArray[i].name)" for "alert(myArray.join)". I got undefined for each item. The other weird thing is that if I do "alert(myArray.length)" it says there are 14 items in the array, but there should be 8....

So using your code, I modified the function to this, and it seems to work:

Code:

function DisplayTags() {
  var obj = document.getElementById('form_id');
  var _input = obj.getElementsByTagName('input');
  var _select = obj.getElementsByTagName('select');
  var myArray = new Array();

	for(var i=0; i<_input.length; i++){ myArray[myArray.length] = _input[i];}
	for(i=0; i<_select.length; i++){ myArray[myArray.length] = _select[i]; }
	
	for(i=0; i<myArray.length; i++){
		alert("name: "+myArray[i].name+"\nitem "+(i+1)+" of "+myArray.length);	
	}
}

It's the same basic idea, but for some reason, when addressing the array with the index value, it seems to work.

Thanks for your help. I was hoping for a cleaner solution, like using a method of the Array object, but I can't seem to figure it out.

Thanks again!
Brian

**JMRKER** · 05-15-2009 11:17 AM

You're most welcome.
I guess we'll need the help of some more experienced forum members to explain what's happening

,
but so long as it works for you, I'm happy ...

Good Luck!

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