what am i doing wrong?
i want to populate the country list according to the category
i want to switch in case CategoryID=1 and CatID2=2 in the country_list table
then call on "alpha" to populate the dropdown list via ajax
the form:
Code: <!--
The following file was created by Iwo Kadziela at TECHROAM: http://www.techroam.com
This file uses MySql database and AJAX to create autopopulating dynamic select boxes.
<html>
<head>
<script src="request.js"></script>
<script type="text/javascript">
function getfirst(dd1)
{
clearSelect = document.getElementById('thirdbox');
clearSelect.options.length = 1;
//above code clears the third select box when re-clicking the first select box
var idx = dd1.selectedIndex;
var first = dd1[idx].value;
var par = document.forms["theform"];
var parelmts = par.elements;
var prezsel = parelmts["secondbox"];
if (first != "Select Something One")
{
Http.get({
url: "./alpha.php?alpha=" + first + "%",
callback: fillPrez,
cache: Http.Cache.Get
}, [prezsel]);
}
}
function getsecond(dd1)
{
var idx = dd1.selectedIndex;
var second = dd1[idx].text;
var par = document.forms["theform"];
var parelmts = par.elements;
var prezsel = parelmts["thirdbox"];
if (second != "Select Something Two")
{
Http.get({
url: "./beta.php?state=" + second,
callback: fillPrez,
cache: Http.Cache.Get
}, [prezsel]);
}
}
function fillPrez(xmlreply, prezelmt)
{
if (xmlreply.status == Http.Status.OK)
{
var prezresponse = xmlreply.responseText;
var prezar = prezresponse.split("|");
prezelmt.length = 1;
prezelmt.length = prezar.length;
for (o=1; o < prezar.length; o++)
{
prezelmt[o].text = prezar[o];
prezelmt[o].value = prezar[o];
}
}
else
{
alert("Cannot handle the AJAX call.");
}
}
function getthird(dd1,dd2) {
var idx = dd1.selectedIndex;
var third = dd1[idx].text;
if (third != "Select The Final")
{
Http.get({
url: "./final.php?state=" + dd2 + "&city=" + third,
callback: getit,
cache: Http.Cache.Get
});
}
}
function getit(xmlreply){
if (xmlreply.status == Http.Status.OK)
{
document.getElementById("final").innerHTML= xmlreply.responseText;
}
else
{
alert("Cannot handle the AJAX call.");
}
}
</script>
<style>
body {
width:800px;
margin: 0px auto;
text-align:center;
}
p {display:inline;margin:20px;}
div {clear:both;width:800px;border:1px solid #999;padding:20px;margin:20px 0px 20px 0px;}
</style>
</head>
<body>
<form name="theform">
<p>
<select name="firstbox" size="13" style="position: relative; width: 215px;" onChange="getfirst(this);">
<option value="">Select Something One</option>
<?php
do {
?>
<option value="<?php echo $row_RsCategories['CategoryID']?>"><?php echo $row_RsCategories['CategoryName']?></option>
<?php
} while ($row_RsCategories = mysql_fetch_assoc($RsCategories));
$rows = mysql_num_rows($RsCategories);
if($rows > 0) {
mysql_data_seek($RsCategories, 0);
$row_RsCategories = mysql_fetch_assoc($RsCategories);
}
?>
</select>
</p>
<p>
<select name="secondbox" size="13" style="position: relative; width: 215px;" onChange="getsecond(this);">
<option>Select Something Two</option>
</select>
</p>
<p>
<select id="thirdbox" name="thirdbox" size="13" style="position: relative; width: 215px;" onChange="getthird(this, document.theform.secondbox.options[document.theform.secondbox.selectedIndex].value);">
<option>Select The Final</option>
</select>
</p>
</form>
<center>THE FINAL CONTENT PULLED FROM DATABASE</center>
<div id="final"></div>
</body>
</html>
<?php
mysql_free_result($RsCategories);
?>
alpha.php
Code: </php
$dbconnection = mysql_connect ("localhost", "root", "") or die ('I cannot connect to the database because: ' . mysql_error());
mysql_select_db ("international", $dbconnection) or die("Couldn't open database: ".mysql_error());
$alpha=$_GET['alpha'];
switch ($alpha)
{
case 1:
$result = mysql_query("SELECT CountryName from country_list WHERE CategoryID = '".$_GET['alpha']."'");
break;
case 2:
$result = mysql_query("SELECT CountryName from country_list WHERE CatID2 = '".$_GET['alpha']."'");
break;
while ($row = mysql_fetch_array($result)) {
echo "|" . $row['CountryName'];
}
?>
PHP page is blank
Hi Everyone,I have a site in which I am able to open the first PHP webpage in my browser but when I start navigating the site, another webpage doesn't open. The page is totally blank for this webpage.
SAP Management Console is blank
Dear All ,I am facing problem in my des sever suddely in SAP Management Console is blank and there is no tree struce of sidand oracle is up and SAPOSCOL services is running but SAPDES_00 services
Google Map
I have done Google Map Integration for one my project. But for that we need the "latitude and longitude" for this. How we will find the these values using PHP 5.1.6. In 5.2, we can do it
fwrite error
Hi All,Does anyone know what is causing the error in this code?Code: <?$errorLog_FilePath = "error_log.txt";$handle = @fopen($errorLog_FilePath,'a+');$strError =
DirectoryIterator and Hacked Website
Hi Everyone.My problem:Some one has been sneaking in to my website hidden Iframes. I've changed passwords & reset permissions to read, but then the permissions get changed to include write and
Hotlinking Picasa as the image folder of a website
Hi there PHP freaks, I would like to create a private album in Picasa to use it as the image folder for my website, I want this album private so images can only be accessd from my website, so I guess
getting rid of quotes in strings
sick of trying to deal with them in multiple ways (entering in and taking from database, echoing, echoing within a text input, etc) so i would just like to get rid of them.turn: bob's houseinto: bobs
Get content from table into a list, without repeating.
Alright this is kinda an odd thing, so I need some help.I have a table "quote" with the three columns `id`, `quote`, and `author`.Now I am wondering if there is anyway that I can output all
parse error
Parse error: syntax error, unexpected '[', expecting ')' in /Users/admin/Sites/phptest/array.php on line 5 Code: <?phpfunction shippingPrice($zone,$weight){ $ground = array(2 => array
Need help urgant
why down my code ony return one item instead of all the items selected. $arr =