convert this code to drop down menu?

FiveFlat

This following code lists all the names and makes a link out of each horse. I want to convert this to a drop down menu that automatically goes to the new page without using a "go" button. Can anyone help me out with this please?

<?php
require('commonvars.php');
mysql_connect(localhost,$dbuser,$dbpass);
@mysql_select_db($database) or die ("Unable to connect!");
$query="SELECT * FROM horse";
$result=mysql_query($query);

echo ("<b><font size=4>SELECTION</font></b><br><br>");
$num=mysql_numrows($result);
$i=0;
while ($i < $num) {
$reg_num=mysql_result($result,$i,"reg_num");
$reg_name=mysql_result($result,$i,"reg_name");
$nickname=mysql_result($result,$i,"nickname");
$bdate=mysql_result($result,$i,"bdate");
$sex=mysql_result($result,$i,"sex");
$sire=mysql_result($result,$i,"sire");
$dam=mysql_result($result,$i,"dam");

echo ("<a href=$PHP_SELF?pedigree=$reg_num><b>$reg_name</b></a><br>");
++$i;
}
?>

vaaaska

You're close...

Load up the form stuff and include it with your loop...echo it out...

echo "<select name='asdf'>\n"

then the loop...

echo "<option value='$value'>$name</option>

and of course echo "</select>" at the end...

THEN...you'll need to get some javascript to turn it into a jump menu, but that's pretty easy to do as well...but this is a php forum afterall...

good luck...v

FiveFlat

thanks! I got the menu looking correct. I've decided to just use a GO button for now - but it doesn't work. Do you see anything wrong with my code?

<?php
require('commonvars.php');
mysql_connect(localhost,$dbuser,$dbpass);
@mysql_select_db($database) or die ("Unable to connect!");
$query="SELECT * FROM horse";
$result=mysql_query($query);

echo ("<b><font size=4>SELECTION</font></b><br><br>");
$num=mysql_numrows($result);
echo ("<select name='select'>");
$i=0;
while ($i < $num) {
$reg_num=mysql_result($result,$i,"reg_num");
$reg_name=mysql_result($result,$i,"reg_name");
$nickname=mysql_result($result,$i,"nickname");
$bdate=mysql_result($result,$i,"bdate");
$sex=mysql_result($result,$i,"sex");
$sire=mysql_result($result,$i,"sire");
$dam=mysql_result($result,$i,"dam");
echo ("<option value='$reg_num'>$reg_name</option>");
++$i;
}
echo ("</select><a href=$PHP_SELF?pedigree=$select><img src=images/go_button.gif></a>");
?>

the problem is at the end of the options I have my link. the $select var isn't working. Did do it completely wrong? Can you tell I am not a pro?

FiveFlat

after "googling" 'how to drop down menu' It appears that drop down menus and javascript go hand in hand? You can't make a drop down menu without incorporating java eh?

vaaaska

no, they don't necessarily go hand in hand...when you say "drop down" i believe you really mean "list menu". i believe you want a form element and not something else...

as far as creating a "jump menu" here is a link to some code you can adapt into your page...

http://javascript.internet.com/navigation/jump-menu.html

😉