Thanks for the help it worked. Knew it was very obvious!
On Feb 2, 8:46 am, John Andersen <j.andersen...@gmail.com> wrote:
> This groups is for the CakePHP framework, which is somewhat more than
> regular PHP, so I suggest you try the comp.lang.php group for future
> questions, as they will be able better to help you!
>
> On the other hand, you need to make the following change to your code:
>
> [code]
> $sType = $_GET['type'];
> while($type = mysql_fetch_array($xtype)) {
> $selected = '';
> if ( $type['type'] == $sType ) {
> $selected = 'selected="selected"';
> }
> echo "<option value='$type[type]' $selected>$type[type]</
> option>"."<BR>";}
>
> [/code]
>
> This will make the first select show the chosen value, after the page
> is reloaded. Note - code not tested!
>
> Observe that I am not checking for whether or not the $_GET parameter
> exists and not doing anything against code injection - that is for
> your future work, not now just for making this work!
> Enjoy,
> John
>
> On Feb 1, 10:15 pm, "Marc A." <questionm...@gmail.com> wrote:
>
> > Hi,
>
> > I'm not very experienced with PHP and I've been told this is a good
> > place to ask for help.
>
> > I have dyanamic dropdowns in a form taken from a DB. Meaning the first
> > dropdowns is populated by items in the DB and once selected it reloads
> > the page with javascript and populates the second dropdown based on
> > the first one.
>
> > Ive gotten this to work, however my problem is that once it reloads it
> > populates the second dropdown but the first dropdown goes back to the
> > defaulted select option. I want it to stay on the selected one that
> > populate the second dropdown.
>
> > Here is the javascript ive used:
>
> > <SCRIPT language=JavaScript>
> > function reload(form){
> > var val=form.type.options[form.type.options.selectedIndex].value;
> > self.location='dropdown.php?type=' + val ;}
>
> > </script>
>
> > and heres my PHP code:
>
> > <?
> > include("dbinfo.inc.php");
> > mysql_connect(localhost,$username,$password);
> > @mysql_select_db($database) or die( "Unable to select database");
>
> > $xtype=mysql_query("SELECT DISTINCT type FROM prod");
>
> > echo "<form method=post name=f1 action='dropresult.php'>";
>
> > ////////// Starting of first drop downlist /////////
> > echo "<select name='type' onchange=\"reload(this.form)\">";
> > while($type = mysql_fetch_array($xtype)) {
> > echo "<option value='$type[type]'>$type[type]</option>"."<BR>";
> > }
> > echo "</select>";
>
> > echo $sType;
>
> > ////////// Starting of the second drop downlist /////////
> > $xquality = mysql_query("SELECT DISTINCT quality FROM prod WHERE
> > type='".$sType."'");
>
> > echo "<select name='quality'><option value=''>Select one</option>";
> > while($qual = mysql_fetch_array($xquality)) {
> > echo "<option value='$qual[quality]'>$qual[quality]</option>";
> > }
> > echo "</select>";
>
> > echo "<input type=submit value=Submit>";
> > echo "</form>";
> > ?>
>
> > I have a feeling the solution might be very simple however it seems to
> > escape me immensly.
>
> > Any help is very much appreciated.
>
> > Thanks in advance
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