Function can't be repeated????

[expert_21]

Hi, I am trying to repeat a list menu in a form, but nothing shows up. returnDCatDropDown returns the HTML for the dropdown. Please assists.

function returnDCatDropDown($name='dCat_{counter}',$initialselect=0)
{
	global $siteDB;
	$dropDown .= "<select name=$name id=$name>";
	$siteDB->query('SELECT id, name FROM ' .DB_PREFIX. '_download_cat;');
	$dropDown .= "<option value=\"0\">Main Category</option>";
	while ($array = mysql_fetch_row($siteDB->results))
	{
		if ($array[0]==$initialselect)
		{
			$dropDown .= "<option value=\"$array[0]\" selected=\"selected\">$array[1]</option>";
		}
		else
		{
			$dropDown .= "<option value=\"$array[0]\">$array[1]</option>";
		}
	}
	$dropDown .= '</select>';
	return $dropDown;
}

$modifyForm='<input name="deleteCategory[]" type="checkbox" id="deleteCategory" value="{counter}">';
$modifyForm .= 'Category: 
  <input name="category_{counter}" type="text" id="category_{counter}" value="{value}">
  <br>
  Parent:{catDropDown}<br>Description:<textarea name="description_{counter}"></textarea><br></br>';
		$siteDB->query('SELECT id, name, parent FROM ' .DB_PREFIX. '_download_cat;');
		$dropdown=returnDCatDropDown();
		while ($array = mysql_fetch_row($siteDB->results))
		{	
			$newModifyForm= str_replace('{catDropDown}',$dropdown,$modifyForm);
			$newModifyForm= str_replace('{counter}',$array[0],$newModifyForm); // replace with cat_id
			$newModifyForm= str_replace('{value}',$array[1],$newModifyForm); // replace with cat name
			print $newModifyForm;
		}

[sid]

do you get an error? or whats the exact problem?

to give it a start: this here

function  returnDCatDropDown($name='dCat_{counter}',$initial select=0)

is no valid php.

[expert_21]

hi sid, thanks for the reply. I didn't get any error, it just returns nothing when it is suppose to return a list box. I think this is a valid php:

function returnDCatDropDown($name='dCat_{counter}',$initialselect=0)

the whole php is in one line, it just got cut off while pasting here 🙂 there's nothing wrong with right? (i hope)

Now this is the part that confuses me.

this is the return drop down function:

function returnDCatDropDown($name='dCat_{counter}',$initialselect=0)
{
	global $siteDB;
	static $ran, $dropDown;
	if (!$ran)
	{
		$dropDown .= "<select name=$name id=$name>";
		$dropDown .= "<option value=\"0\">Main Category</option>";
		$siteDB->query('SELECT id, name FROM ' .DB_PREFIX. '_download_cat;');
		while ($array = mysql_fetch_row($siteDB->results))
		{
			$dropDown .= "<option value=\"$array[0]\">$array[1]</option>";	
		}
		$dropDown .= '</select>'; // dropdown created, can begin changing selected option
		$ran=TRUE;
		if ($initialselect !=0)
		{
			$newDropDown= str_replace('value="'.$initialselect.'"','value="'.$initialselect.'"'.' selected="selected"', $dropDown);
			return $newDropDown; // modified, selected option
		}
		else
			return $dropDown; // unmodified
	}
	else
	{
		if ($initialselect !=0)
		{
			$newDropDown= str_replace('value="'.$initialselect.'"','value="'.$initialselect.'"'.' selected="selected"', $dropDown);
			return $newDropDown; // modified, selected option
		}
		else
			return $dropDown; // unmodified

}
}

and this is the code that calls the function:

		$dropdown=returnDCatDropDown();
		$siteDB->query('SELECT id, name, parent FROM ' .DB_PREFIX. '_download_cat;');
		while ($array = mysql_fetch_row($siteDB->results))
		{	
			$dropdown=returnDCatDropDown('parent_id_{counter}',$array[2]);
			$newModifyForm= str_replace('{counter}',$array[0],$modifyForm); // replace with cat_id
			$newModifyForm= str_replace('{catDropDown}',$dropdown,$newModifyForm); // replace with parent_id
			$newModifyForm= str_replace('{value}',$array[1],$newModifyForm); // replace with cat name
			print $newModifyForm;
		}

notice the first line?: $dropdown=returnDCatDropDown(); Without the line in bold, the script returns nothing, but with the line, it works fine (drop down list returned). I am EXTREMELY confused 😕 can you shed some light on this?