php if in an option tag

Question

hi can anyone help me with this? i really had a hard time putting my if statement in this code

<select name="date_opened_year">
<?php 
define('DOB_YEAR_START', 1962);
$current_year = date('Y');

for ($count = $current_year; $count >= DOB_YEAR_START; $count--)
{
    echo "<option value='{$count}'>{$count}</option>";
}
?>
</select>

and i have here a code where i had put my if statement

<select name="date_opened_year">
<?php 
define('DOB_YEAR_START', 1962);
$current_year = date('Y');

for ($count = $current_year; $count >= DOB_YEAR_START; $count--)
{
    echo "<option value='{$count}' if($yy == $count) echo 'selected'>{$count}</option>";
}
?>
</select>

$yy = 2003 well it comes from a query in the database and when i look at the code it gives me something like this

<option 'selected'="" echo="" 2012)="" if(2003="=" value="2012">2012</option>

why is that? can anyone revise the code? thanks

Answer 1

try this:

$selectStr = ($yy == $count) ? 'selected': '';
echo "<option value='".$count."' ".$selectStr.">".$count."</option>";

Best to break it up.

You've got your if inside a string creation double quote.

Answer 2

do this:

if($yy == $count){ 
   echo "<option value='{$count}' selected='selected'>{$count}</option>";
}

Answer 3

Echo will not evaluate if statement. Try @DCoder solution with printf or this:

echo "<option value='{$count}'";
if($yy == $count) echo 'selected';
echo ">{$count}</option>";