I have the following jQuery code:

```
$('#btnAdd').click(function() {
var i = ($('#stuff >tbody >tr').length)+1;
$('#stuff >tbody >tr:last').clone(true).find("input,select").each(function(){
$(this).attr({
'name': function(_, name){
   return name + i;
},
'value': ''
});}).end().insertAfter('#stuff >tbody>tr:last').show();

```

Which produces the following DOM:

```
<tr>
<td class="td_6">
<select name="name1">
<option selected="" value="Bricklayers">Bricklayers</option>
<option value="Sheet Metal Workers">Sheet Metal Workers</option>
<option value="Sprinkler Fitters">Sprinkler Fitters</option>
</select>
<td>
<td class="td_6">
<input name="V_union" value="1" type="radio">Yes<input name="V_union" value="0" type="radio" checked="checked">No
</td>
<td class="td_6">
<input name="V_picketed" value="1" type="radio">Yes<input name="V_picketed" value="0" type="radio" checked="checked">No
</td>
<td>
</td>
</tr>

<tr style="">
<td class="td_6">
<select name="name13" value="">
<option selected="" value="Bricklayers">Bricklayers</option>
<option value="Sheet Metal Workers">Sheet Metal Workers</option>
<option value="Sprinkler Fitters">Sprinkler Fitters</option>
</select>
</td>
<td class="td_6">
<input name="V_union3" value="" type="radio">Yes<input name="V_union3" value="" type="radio" checked="checked">No
</td>
<td class="td_6">
<input name="V_picketed3" value="" type="radio">Yes<input name="V_picketed3" value="" type="radio" checked="checked">No
</td>
<td>
</td>
</tr>

```

Here is the form tag:

```
<form method="post" action="view_job.php" name="updateViolations" enctype="multipart/form-data">

```

The dynamically entered rows are within the tags.

When I do a var\_dump on $\_POST, I am not seeing the new dynamically generated form inputs in the $\_POST variable.

I have checked several other questions, and it appears that I am doing everything right, but I'm lost.