drop down changes from other drop down with calculation. Jscript / Ajax? - php

Question

I have a Jscript Query.
I have done a bit of reading and found out that AJAX is just a side server for a lfash script that can be used on Linux with php. (please correct me if I have interperated that wrong)
I have no knowledge on how scripts work so this is new, I have tried a couple of different tries but no luck.
I have one drop down box (Box1) (populated from Database)
I have another box (Box2) for a calculation to insert into my database for other uses on ohter parts of hte site.
I need the Box2 to change the figure when someone changes Box1 dropdown before hitting the submit button.
I think because I have the calcualtion this is getting me stuck... Code is as below... Can someone please help me figure out (I think I need some form of Script to do this.) the answer...
Box1
<td><p>selection 1</p>
<select id="t1_type" name="t1_type">
<?php $result = mysql_query("SELECT * FROM `t2` ORDER BY t2_value");
while($valuerow = mysql_fetch_array($result)){
echo '<option value="'.$valuerow['t2_name'].'">'.$valuerow['t2_name'].'</option>'; } ?>
Box2
<input name="t1_value" id="t1_value" value="
<?php
$var1 = $row_value['t2_value'];
$var2 = $row_dropdown['t1_number'];
$total = round ($var2 * $var1);
echo "" . $total . "";
?>" />
I hope this is all the code you need, (Let me know if more required)
What it needs to do is show new calculation whenever someone changes the box1 option BEFORE the submit button is clicked, so it submits the correct calculation to the database for future use.
I think it would need pretty much "t2_value" from box2 to change when ever "t2_name changed from box1.
And once again the best link to learn about the solution. (Learnt about Joins now from my last question!! Almost a intermediate user. ;-) )

Answer 1

Edit :
I saw that your second box was a textbox I believe, if thats the problem then you should do something like this
<select id="t1_type" name="t1_type" onchange="change(this);">
<?php
$result = mysql_query("SELECT * FROM `t2` ORDER BY t2_value");
while($valuerow = mysql_fetch_array($result))
{
echo '<option value="'.$valuerow['t2_name'].'">'.$valuerow['t2_name'].'</option>';
}
?>
</select>
This defines your <select> box like you did in your question.
Add an onChange event into your first <select> and then create a function to handle to onChange event. An onChange event fires whenever the user changes the item in the <select> element.
Javascript :
( put this part of the code above the </head> )
<script language="javascript" type="text/javascript">
function change(element)
{
// do here whatever you want
// you can change the value of the <input> box with :
// document.getElementById(element.id).value = your_value
// If you want to see if this part works, then try adding this :
// alert("It works!");
// If you want to get the text of the item which has been selected in Box1 use :
// $("#t1_type option:selected").text();
}
</script>
Note: because PHP is server side, you can't update your Box2 dynamically without a page refresh, Javascript however is Client Side and CAN do this.
Note : the $("#t1_type option:selected").text(); code requires you to include the jQuery library into your script. Be sure to convert this variable to a float, int or double if you want to calculate with it, else the outcome will give NaN (Not a Number)
Tutorial on including jQuery Libary :
http://docs.jquery.com/Tutorials:Getting_Started_with_jQuery
If you're new to JavaScript, you should try some tutorials. The ones at w3Schools.com helpt me alot, but some people say they're not always correct, but anyhow, read some stuff about Javascript to actually know what you are doing, instead of copypasting code :)