[SOLVED] Help auto-selecting field

[msaz87]

Hey all,

 

I'm having difficulties getting a drop down list to auto select the current date from a MySQL DB. The code below shows the the date getting pulled then formatted and my attempt at getting it to select.

 

						while($row = mysql_fetch_array($tournament_details_results)){ 

						$start	= $row['date'];
						$days	= $row['length'];

					}

					$fix_date	= strtotime($start);

					$date		= date('M j, Y',$fix_date);

					?>

					Date: <br/>
					<select name="date">
					<?php for ($e = 0; $e < $days; $e++, $t++) { ?>

							<?php 		

								$s	= date('M j, Y', strtotime($date)+(86400*$t));
								$sv	= date('Y-m-d', strtotime($s));

							?>							

						<option <? if ($date_selection_mod == ($sv)); echo "selected"; ?> value="<?php echo($sv); ?>">

							<?php echo($s); ?>

						</option>

					<?php } ?>
					</select>

 

When setting up the DB, the user inputs how long the event is, in terms of days, then gives a start date and so the $e portion deals with making all the dates for it automatically, though it only stores the start one.

 

Any ideas? Thanks very much!

[ngreenwood6]

well I see a bunch of problems with your code. First of all you are looping through all of the results and replacing the old value with the new one. So $start & $days will only have one value. Second of all you could simply loop through the results and display it all at one time. Try this:

 

Date: <br/>
<select name="date">
  <?php while($row = mysql_fetch_array($tournament_details_results)){                    
  	$start   = $row['date'];
    $days   = $row['length'];
    $fix_date   = strtotime($start);
    $date      = date('M j, Y',$fix_date);
    $s   = date('M j, Y', strtotime($date)+(86400*$t));
    sv   = date('Y-m-d', strtotime($s));
  ?>
  <option <?php if ($date_selection_mod == ($sv)); echo "selected"; ?> value="<?php echo($sv); ?>"> <?php echo($s); ?> </option>
  <?php } ?>
</select>

[msaz87]

well I see a bunch of problems with your code. First of all you are looping through all of the results and replacing the old value with the new one. So $start & $days will only have one value. Second of all you could simply loop through the results and display it all at one time. Try this:

 

Date: <br/>
<select name="date">
  <?php while($row = mysql_fetch_array($tournament_details_results)){                    
  	$start   = $row['date'];
    $days   = $row['length'];
    $fix_date   = strtotime($start);
    $date      = date('M j, Y',$fix_date);
    $s   = date('M j, Y', strtotime($date)+(86400*$t));
    sv   = date('Y-m-d', strtotime($s));
  ?>
  <option <?php if ($date_selection_mod == ($sv)); echo "selected"; ?> value="<?php echo($sv); ?>"> <?php echo($s); ?> </option>
  <?php } ?>
</select>

 

What happens with the code you suggested is it only displays one date (and it's actually a date earlier than the one retrieved from the DB).

 

Let me try to explain it in context a bit more...

 

This page is used to edit a schedule of an event. The event is set up in a previous page where they enter a start date, say 2009-11-20 and then say how many days it lasts. So lets say it lasts 4 days, the only two things entered into the DB are the 2009-11-20 and the length...

 

On that <select> field, the original code pulled the start date, then used the loop to generate the next three days as options.. IE: so you can set up a schedule for each day of the tournament.

 

Since this page is used to edit these, I'm looking to make it auto-select the date entered into the scheduling DB (so if when setting it up, they picked 2009-11-22, it would default to that).

 

Not sure if that helps clear anything up... if I'm not giving enough code to make it clear, please let me know.

 

Thanks for the help!