drop down box

[New_Coder]

i have this menu

<select name="month">
<option value="2003">2003
<option value="2004">2004
<option value="2005">2005
<option value="2006">2006
<option value="2007">2007
<option value="2008">2008
<option value="2009">2009
<option value="2010">2010
<option value="2011">2011
<option value="2012">2012
<option value="2013">2013
<option value="2014">2014
</select>

now how to show selected defualt date ?? I need a way for the code to automatically select 2009 for 2009, but in 2005, 2005, etc.

[NogDog]

<select name="year">
<?php
$thisYear = date('Y');
for($yr = 2003; $yr <= 2014; $yr++)
{
   printf(
      "<option value='&#37;04d'%s>%04d</option>\n",
      $yr,
      ($yr == $thisYear) ? " selected='selected'" : '',
      $yr
   );
}
?>
</select>

[New_Coder]

NogDog;10905262 wrote:

<select name="year">
<?php
$thisYear = date('Y');
for($yr = 2003; $yr <= 2014; $yr++)
{
   printf(
      "<option value='%04d'%s>%04d</option>\n",
      $yr,
      ($yr == $thisYear) ? " selected='selected'" : '',
      $yr
   );
}
?>
</select>

yes ! this worked in php+html page . but i have .TPL file and not use printf and import html code . <select> and <option> in my template file (.TPL) . for me all php code = $test; and $test = {test} so in template {test} display . now how to ?

[New_Coder]

any idea for help ?

[NogDog]

New Coder;10905332 wrote:

any idea for help ?

Since I don't know your template system, no.

Maybe you can use the above method in the main script, but assign it to a variable instead of printing it (using sprintf() instead of printf() and appending it to a variable via ".=" in the loop), then send that to your template in whatever method it uses.

[New_Coder]

OK ! fix with template . now how to all code or sprintf = $test ?? i need for display to $test . e.g = echo $test ! ( diplay <option> menu ) . or any method for display $test ( only ) . your code display menu but i need to only display with $test . this is Fix . how to ?

[New_Coder]

any help please !