Hello,
I searched Cake PHP for "DOM" and got 4 pages of results, none of which
appeared to address this question.
I am having a problem with code adapted from the first edition of
"Programming PHP." This code dynamically generates checkboxes or radio
buttons that retain their settings when the form is submitted without
passing validation.
The problem is that these dynamically generated form controls almost
never appear in the DOM when the page loads. I am only aware of one
instance with Firefox 2 where I was able to access these form controls
with DHTML. My code example appears below.
<?php
function make_radio ($name, $query, $options) {
$i = 0;
foreach ($options as $value => $label) {
printf('<label><input type="radio" name="%s[]" id="%s_%s"
value="%s" ',
$name, $name, $i, $value);
if (in_array($value, $query)) {echo "checked ";}
echo "/> $label</label><br />\n";
$i++;
}
}
if (! is_array($own_files)) { $own_files = array(); }
$preference1 = array(
'Yes' => 'Yes',
'No' => 'No',
);
make_radio (own_files, $own_files, $preference1);
?>
As I said before, there was one instance where these form controls did
make it into the DOM. Otherwise, the following examples test as
undefined.
<script type="text/javascript">
document.forms[0].own_files[0];
document.forms[0].own_files[1];
document.getElementByID('own_files_0');
document.getElementByID('own_files_1');
</script>
Does anyone know how to get dynamically generated form controls into
the DOM on a reliable basis? I have tested the above JavaScript in
Firefox 2, IE 7, and Opera 9 for Windows XP. Thank you to all who can
shed light on this problem.
--
Mountain Man
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