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PHP: 10 years in drop-down options

Posted on 2009-03-29
7
213 Views
Last Modified: 2012-05-06
I want to use PHP to fill in drop-down options with the next 10 years.
<select name="ExpYear">
				<option value="2009">2009</option>
				<option value="2010">2010</option>
				<option value="2011">2011</option>
				<option value="2012">2012</option>
				<option value="2013">2013</option>
				<option value="2014">2014</option>
				<option value="2015">2015</option>
				<option value="2016">2016</option>
				<option value="2017">2017</option>
				<option value="2018">2018</option>
			</select>

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Question by:hankknight
  • 3
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7 Comments
 
LVL 3

Assisted Solution

by:Ruurtjan
Ruurtjan earned 200 total points
ID: 24012606
Is this what you're looking for?
<?
echo '<select name="ExpYear">';
for($year = date("Y"); $year < date("Y")+10;$year++) {
	echo sprintf("<option value=\"%d\">%d</option>\n", $year, $year);
}
echo '</select>';
?>

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LVL 39

Accepted Solution

by:
Roger Baklund earned 300 total points
ID: 24012614
Try this:
<select name="ExpYear">
<?php
$stop = (int)date('Y') + 10;
for($y = date('Y'); $y < $stop; $y++)
  echo '<option value="'.$y.'">'.$y.'</option>';
?>
</select>

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Expert Comment

by:phpretard
ID: 24012686
That worked well.  I searched and did not find that...

With the below format how would I return the value posted and display it as selected?

So if I select 2010 and other parts of the form did not pass validation then then 2010 would stay as the selected value.
<select name="Exp">
 
<?
$stop = (int)date('Y') + 5;
for($y = date('Y'); $y < $stop; $y++)
  echo '<option value="'.$y.'">'.$y.'</option>';
}
?>
 
</select>

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LVL 3

Expert Comment

by:Ruurtjan
ID: 24012700
This should work:
<select name="Exp">
 
<?
$stop = (int)date('Y') + 5;
for($y = date('Y'); $y < $stop; $y++) {
  $selected = '';
  if(isset($_get['Exp']) && $_get['Exp'] == $y)
    $selected = ' SELECTED';
  echo '<option value="'.$y.'"'. $selected .'>'.$y.'</option>';
}
}
?>
 
</select>

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0
 

Expert Comment

by:phpretard
ID: 24013002
That didn't work but I did manage to use your first code an d make it work for this purpose.

Thanks!
$stop = (int)date('Y') + 5;
for($y = date('Y'); $y < $stop; $y++) {
  $selected = '';
  if(isset($_get['LicenseYY']) && $_get['LicenseYY'] == $y)
    $selected = 'SELECTED';
  echo '<option value="'.$y.'"'. $selected .'>'.$y.'</option>';
}
 
// THIS WORKED
 
if (isset($_POST['submit'])){
 
echo"<option>$LicenseYY</option>";
$stop = (int)date('Y') + 5;
for($y = date('Y'); $y < $stop; $y++)
  echo '<option value="'.$y.'">'.$y.'</option>';
}else {
 
$stop = (int)date('Y') + 5;
for($y = date('Y'); $y < $stop; $y++)
  echo '<option value="'.$y.'">'.$y.'</option>';
 
}

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Expert Comment

by:phpretard
ID: 24013010
I don't see a solution button as I am trying to award the points to you.
0
 
LVL 39

Expert Comment

by:Roger Baklund
ID: 24013094
phpretard, the author of this question is "hankknight", only that use can close the question...

Ruurtjan wrote $_get instead of $_GET, the variable name is case sensitive. It should be done like below.

Please split the points when multiple experts have collaborated for a solution.
$stop = (int)date('Y') + 5;
for($y = date('Y'); $y < $stop; $y++) {
  $selected = '';
  if(isset($_GET['LicenseYY']) and ($_GET['LicenseYY'] == $y))
    $selected = ' selected="selected"';
  echo '<option value="'.$y.'"'. $selected .'>'.$y.'</option>';
}

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