This is a useful tutorial

I can't get this to work. I have tried another example where I list each option with an associated id and it works. I need to do this. I have a list populated from a mysql table and I need it to populate fields like this example when a name is selected.

why this script does not work in my project???
this error --> object expected

Hm. I checked the script and everything is fine. Check the html source for:

1. Correct syntax of copied JS function .
2. Correct connection and query to MySQL Data
3. Correct values in <option value=

It's fantastic and amazing, Thanks for showing this amazing magic of ajax

ok,,

Thanks for sharing. The function GetXmlHttpObject() is forgotten? Below is its definition:
function GetXmlHttpObject() {
var xmlhttp=null;
try {
//If the Javascript version is greater than 5.
xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
alert ("You are using Microsoft Internet Explorer.");
return xmlhttp;
} catch (e) {
//If not, then use the older active x object.
try {
//If we are using Internet Explorer.
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
alert ("You are using Microsoft Internet Explorer");
return xmlhttp;
} catch (E) {
//Else we must be using a non-IE browser.
//xmlhttp = false;
//If we are using a non-IE browser, create a javascript instance of the object.
if (typeof XMLHttpRequest != 'undefined') {
xmlhttp = new XMLHttpRequest();
alert ("You are not using Microsoft Internet Explorer");
return xmlhttp;
}

}
}
}

Thanks for the note. We will correct this.

hi there, where i need to paste the PHP server side script ? Thanks

You have to paste server side PHP script in file named ajax.php. This file have to be placed in the same directory where is located your start php file like shown in the source code.

thank you,,,,,,,,,,,,,,,,,,

Do you have an idea how these employ details could be display if there are more than 1 person name John (maybe John Smith, John Doe, John Micheal)? I'm trying to have a drop-down menu that depends on their family name and then will display all records maybe 2, 3 or more with a same family name.

You just have to describe this names (even with same family name) in `name` filed in your database and all other way is the same like here.

I will add some other persons with same first names to show you.

thanks ivory. But could it be 'Adam' or 'Maria' only in the drop-down list? Then the table will show 2 records which are Adam Filshvili & Adam Rodriges with their details. Thanks again.

Yes. You can show only the first name, but if in database have two persons with same first names, how you can distinguish two persons in dropdown menu only by first name?

But, however you want ot show only the first name, and entire name (first, middle and last) is in one database field and they are separated by space:

$first_name=substr($name, 0, strpos($name, ' '));

really nice site

thx :)

I have tried to used your code but unable to get the result what you got. I would appreciate you if you explain breifly about the code you have written.

Any way thanks for your code.

Works like a charm, thanks

Works Perfectly!! :)

okay ra man..no problem about its functionality.

tnx alot

The example is good one. if it with Demo than It is more useful Example.

the scripts working but then the table showing the detail is not appear...how??

I don't understand you.
The demo works fine.

This was exactly what I'd been looking for!! Thanks! Was able to easily adapt it to my own requirements.

Oh and I think Nora wanted to display multiple records at the same time based on the name selected.

I was able to do this easily by putting the <table> inside a while loop:

$query="select * from employ where id=$_GET[employ_id]";
$result=mysql_query($query);
while ($employ=mysql_fetch_array($result)) {

echo "<table border=\"1\">
<tr>
<td>Name:</td>
<td>".$employ[name]."</td>
</tr>
<tr>
<td>Appoint Date:</td>
<td>".$employ[appoint_date]."</td>
</tr>
<tr>
<td>Country:</td>
<td>".$employ[country]."</td>
</tr>
<tr>
<td>City:</td>
<td>".$employ[city]."</td>
</tr>
<tr>
<td>E-mail:</td>
<td>".$employ[email]."</td>
</tr>
<tr>
<td>Phone:</td>
<td>".$employ[phone]."</td>
</tr>
</table>";
}

Hi, the scripts work beautiful, but what if I wanna displaying data drom 3 drop down box/list menu?such as Name, month and city?

Best Regards,

Hi Jack,

In your case you have to prepare second "display_data()" function which have to be handled on onchange event when displaying second drop-down.

This function have to make a GET to another IF in ajax.php server side where needed data for you will be served.

Nice Contents

Works nice except the swedish letters ÅÄÖ will not display OK, What to do?
I can't figure out what to do. I have put "<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
" in both pages.