easy one

[itworks]

hi all!!

i need to make a jump menu with names that are on a table in my MySQL database.

i know how to make a jump menu, and i know how to query the tables entries what i don't know is how to query them to appear in a jump menu.

can anyone give me a hand on this?

thanks in advance.

[prslou]

First write your query, asign it to $query or put it directly into the spot I have $query. Then you put this...

<?
while ($jump_name = mysql_fetch_array($query)) {
$jump_title = $jump_name['column'];
echo '<option>' . $jump_title . '</option>';
}
?>

I assume you want a form drop down. So the complete script might looking something like this...

<form name="form" method="post" action="action_here.php">
<select name="jump_menu">
<?
mysql_connect ('localhost', 'user', 'pass');
mysql_select_db ('database');
while ($jump_name = mysql_fetch_array($query)) {
$jump_title = $jump_name['column'];
echo '<option>' . $jump_title . '</option>';
mysql_close();
}
?>
</select>
</form>

That will priont out <option>title here</option> as many times as it finds it in the database...

And of course you'll need a submit button.

EDIT: I had forgotten one thing, now it's fixed.

[rulinus]

itworks:
this is something very similar (i think) to what you're looking:

#here goes your mysql connection and query
#this is not a jump menu, but a select static menu
<select name="prop" id="prop" style="background-color: #6487DC; color: white">
          <option value="null" selected>select one</option>
          <option value="null" >--------------</option>
          <? if ($result2) { 
while ($r = mysql_fetch_array($result2)) { 
$prop_ID = $r["prop_ID"];
$nombre = $r["nombre"]; 
		  echo "<option value='$prop_ID'>$name</option>\n";
		 } }
		  else{
		  echo "<option value='null' style='background-color: red; color: white'>No names retrieved!</option>";
		  }
		  ?>
#end the menu after the while loop
        </select>

Hope it helps.
rulo