thanx so much i will go for the java approach, let me try first and i will let u knw,, much thanx

am doing it for some company and one of the requirements is produce the alert(so i guess i just have to do it), 1.for the use of javascript, php and the web browser how do i do this? 2.for the use of java/c# you sed i will have to get them install something, which is? thanx a lot for the reply

i am developing a help desk software using php and oracle, when an expert posts a solution to the user who asked it, an alert should pop up at the user's machine to alert them that a solution has been sent end then it should dissappear,i have already produced code to post the solution but i have no idea on how to go with the alerts, please help, thanx

function reload(form)
{
var val=form.cat.options[form.cat.options.selectedIndex].value;
self.location='postquery.php?cat=' + val ;
}

am really new to this i want to know how it functions -which form does it refer to? -what is cat?(an object of?) -options? infact evrthing thanx

by the way when i select a value from a drop down main categoryn the error disappears, why?

sorry guess am just confused,forgot to use the tags, i tried to remove it but when i do that the second dropdown menu stops depending on the first dropdown menu. ok, is there anyway i can generate two dropdown menus one depending on the other the error is Notice: Undefined index: cat in c:\program files\easyphp1-8\www\myhelpdesk\dropdown2.php on line 13 this line reads $cat = $_GET['cat'];//

<?php

//include('connect_db.php');
$quer2 = mysql_query("SELECT DISTINCT typename,probtypeid FROM probtype order by typename");
$cat = $_GET['cat'];//This line is added to take care if your global variable is off

if(!empty($cat))
{
$quer=mysql_query('SELECT DISTINCT `catname` FROM `category` WHERE`probtypeid`= \''.mysql_escape_string($cat).'\' ORDER BY `catname`');
}
else
{
$quer=mysql_query('SELECT DISTINCT `catname` FROM `category` OREER BY `catname`');
}

echo '<form method="post" name="f1" action="postquery.php">';

//Add your form processing page address to action in above line. Example action=dd-check.php
// Starting of first drop downlist
echo '<select name="cat" onchange="reload(this.form)">
<option value="">select problem type</option>';

while($noticia2 = mysql_fetch_assoc($quer2))
{
if($noticia2['probtypeid']==@$cat)
{
echo ' <option selected value="',mysql_escape_string($noticia2['probtypeid']),'">',$noticia2['typename'],'</option>',"\n";
}
else
{
echo ' <option value="',$noticia2['probtypeid'],'">',$noticia2['typename'],'</option>',"\n";
}
}
echo '</select>';

//Starting of second drop downlist
echo '<select name="subcat"><option value="">Select problem category</option>',"\n";

while($noticia = mysql_fetch_array($quer))
{
echo ' <option value="',$noticia['catname'],'">',$noticia['catname'],'</option>',"\n";
}

echo '</select>';

//This will end the second drop down list
echo '</form>';
?>

for the first question: pliz help me, i keep getting the error [i]Notice: Undefined index: cat in c:\program files\easyphp1-8\www\myhelpdesk\postqueryagain.php on line 40[/i] if i remove the line $cat=$_GET['cat']; the second drop down menu doesnt function well thanx guys [code] include('connect_db.php'); $quer2=mysql_query("SELECT DISTINCT typename,probtypeid FROM probtype order by typename"); $cat=$_GET['cat'];//This line is added to take care if your global variable is off if(isset($cat) and strlen($cat) > 0) { $quer=mysql_query("SELECT DISTINCT catname FROM category where probtypeid=$cat order by catname"); } else { $quer=mysql_query("SELECT DISTINCT catname FROM category order by catname"); } echo "

"; //Add your form processing page address to action in above line. Example action=dd-check.php // Starting of first drop downlist echo "select problem type"; while($noticia2 = mysql_fetch_array($quer2)) { if($noticia2['probtypeid']==@$cat) { echo "$noticia2[typename]".""; } else { echo "$noticia2[typename]"; } } echo ""; //Starting of second drop downlist echo "Select problem category"; while($noticia = mysql_fetch_array($quer)) { echo "$noticia[catname]"; } echo ""; //This will end the second drop down list echo "

"; ?> [/code] i want to insert the id of the catname.thanx guys nahelna