becky the code below is spot on (nearly)
stick it on a form that calls itself change the <option name= to <option
value= and when you refresh the form $company will contain the comany name.
you can also use selected in the folowing manner
$regs = pg_Exec($conn, "select region_id, region_name from region
where ".$depdisplay." is not null order by ".$depdisplay." ");
$numregs = pg_numrows($regs);
echo ' Region <select name="reg"
onChange="document.thisForm.submit()">';
echo '<option value=0>Choose a Region</option>';
for ($s=0; $s<$numregs; $s++)
{
$reglist = pg_fetch_array($regs,$s);
echo'<option value="'.$reglist["region_name"].'"';
print ($region == $reglist["region_name"]) ? " SELECTED":"
";
echo '>'.$reglist["region_name"].'</option>';
}
echo'</select>';
> -----Original Message-----
> From: Beckie Pack [mailto:[EMAIL PROTECTED]]
> Sent: 23 April 2001 15:29
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP-DB] calling a column into a popup
>
>
> That's almost it. I want to populate the menu automatically from the
> database without having to manually modify the HTML. I
> believe the below
> will pull the data but it doesn't modify the menu in HTML. I
> tried a few
> things like using the variable name as the same as the
> co_name but the menu
> always comes up blank unless I put an option name in the
> list. it doesn't
> pass the variable for $co_name in the HTML.
>
> thanks,
> boo
>
> > From: "Johannes Janson" <[EMAIL PROTECTED]>
> > Date: Sun, 22 Apr 2001 21:49:36 +0200
> > To: [EMAIL PROTECTED]
> > Subject: Re: [PHP-DB] calling a column into a popup
> >
> > Hi,
> >
> > let me get this right. What I understood is the following:
> >
> > 1. Your admin page where you enter the information of a company
> > 2. A Page with a drop-down list where all the company names
> are listed.
> > (These names should then be linked to an information page with more
> > info?)
> > 3. A page where new companies can be added.
> >
> > well the drop-down list you can do with a simple while and
> a select(html)
> > like
> > this:
> > $result = mysql_query("SELECT companyName FROM companyTable ORDER BY
> > companyName");
> > echo "<select name=company>";
> > while (list($c_name) = mysql_fetch_array($result)) {
> > echo "<option name=$c_name]>$c_name</option>";
> > }
> >
> > If this is what you want (what I'm not sure about) I'm glad
> I understood
> > you. If not supply more detailed information.
> >
> > Cheers
> > Johannes
>
>
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