drop down box

skdzines

I am trying to make a drop down list that pulls a list of shows from a database. I have the list populating correctly. However I want to know how to make it load the page called thumbs.php in a frame called "thumbs" and pass a $showID variable along with it when the user selects it from the list. I don't want them to have to select it and then press enter or some kind of submit button. I want it to automaitcally load the page once it is selected.

I am using a jumpMenu from Dreamweaver MX 2004 to accomplish the drop down menu. This is the code it writes and what I have added to it. Maybe you can take a look at it and tell me what I'm doing wrong.

<?
 <td width='255'><form name='shows' action='http://www.movingtoboise.com/thumbs.php?showID='".$show['showID']."' target='thumbs'>
          <select name='showsMenu' onChange='MM_jumpMenu('parent',this,0)'>'>
		  ");
	?>
	<? 

  $sql = "SELECT showID, date 
		FROM shows ORDER BY showID DESC";
  $res = mysql_query($sql); 
while($show = mysql_fetch_array($res)) {
echo ("
    <option>".$show['date']."</option>
    ");
	}
  ?>
</select>          

   </form>

And this is the javascript it adds as well:

Any ideas as to how I can do this. You can see the page by going to www.movingtoboise.com and then selecting "pics" from the navigation.

Thanks for your help.

Frederick

Your target of the form must match the frame name

This does the trick
onchange="this.form.submit();"

<form action="..." method="post" target="framename">
    <select name="name" size="1" onchange="this.form.submit();">
        <option value="">Choose</option>
        <option value="value">Something</option>
    </select>
</form>

skdzines

Thanks Frederick. That does what I want it to however, now the variable is not being passed to the thumbs page when I select a show from the drop down list. This is my code:

<?
 <td width='255'><form name='shows' method='post' action='http://www.movingtoboise.com/thumbs.php?showID='".$show['showID']."' target='thumbs'>
          <select name='showsMenu' onChange='this.form.submit();'>'>
		  <option selected value=0>Select a Show</option>
		  "); 
	?>
	<? 

  $sql = "SELECT showID, date 
		FROM shows ORDER BY showID DESC";
  $res = mysql_query($sql); 
while($show = mysql_fetch_array($res)) {
echo ("

<option value=".$show['showID'].">".$show['date']."</option> 
");
}
  ?>
</select>          

   </form>

I then have this at the top of my thumbs.php page:

$showID = $_POST['showID'];

Any ideas? Thanks again for you help.

Frederick

I see one issue you may be having, common error.

When you use single quotes in your code for using in your elements of your HTML, variable won't render.

Don't mean to be rude, but using single quotes is the laxy way of not having to place backslash before every double quote you want to cancel out.

When dealing with functions and processing code, you can use single quotes, but when it comes to printing out your results within html with variables inside, you want to use doubles quotes.
(Tip: view your source from your browser to see if the values show) Right-Click > View Source

Also, always use <?php and not <? to start your code.

Example:

<?php
echo ("<option value='$variable'>".$show['date']."</option>");
?>

Try using:

<?php
echo "<option value=\"$variable\">Text</option>\n";
?>

Summary:

<?php
echo "<td width=\"255\"><form name=\"shows\" method=\"post\" action=\"thumbs.php\" target=\"thumbs\">\n";
echo "<select name=\"showsMenu\" onChange=\"this.form.submit();\">\n"; 
echo "<option selected value=\"\">Select a Show</option>\n";

$sql = "SELECT showID, date FROM shows ORDER BY showID DESC";
$res = mysql_query($sql);
while($show = mysql_fetch_array($res)) { 
    echo "<option value=\"".$show['showID']."\">".$show['date']."</option>\n"; 
}
echo "</select>\n";
echo "</form>\n";
?>

Now, the next question is, are you pointing your form back to the same page as the drop menu because the page with the ifame in it must be the action address and target the ifame name.

skdzines

I really appreciate your help Frederick. But I really don't understand why my value is not being passed. As for using the single quotes rather than \", single quotes have always worked for me in the past.

Basically I had everything working before I switched to listing the show dates in a drop down box. I was oringinally doing this:


$sql = "SELECT showID, date 
		FROM shows ORDER BY showID DESC";
  $res = mysql_query($sql); 
while($show = mysql_fetch_array($res)) {

echo " <a href='http://www.movingtobosie.com/thumbs.php?showID='".$show['showID']."' target='thumbs'>'".$show['date']."'</a>";
}

This would work fine and load the correct thumbnail images.

Once I have changed it to the drop down list it simply does not want to send the variable with the onChange=this.form.submit();

all of these pages, shows.php, thumbs.php and imageDisplay.php are all set inside of the frameset page called pics.php. shows.php is in the frame named "shows", thumbs.php is in the frame "thumbs" and imageDisplay is in the frame named "display".

This is why I have the form action set to this:
action='http://www.movingtoboise.com/thumbs.php' target='thumbs'

I don't want the shows.php page with the drop down list to refresh, I simply want it to tell the thumbs.php page to reload and send the showID variable so the sql query on it can query the db and retrieve the correct thumbnails according to that ID.

Like I said it works doing the links but not with the drop down list which is what the customer wants.

skdzines

got it working now. All I needed to change was the form method. I changed it from post to get and all works.

Thanks again for your help.

Installer

Don't mean to be contentious, but using single quotes is considered by many,
not just me, as a way to make code readable.