Solved: Php+drop Down Menu

Mr.LLB · 18-Jun-2008, 08:34 PM #1

Hello all,

I have a databse that lists all the product NAME and CODES.
I also have a form in which a drop down menu exists. In the drop down menu, all the codes of the products are populated. What I want to do is once the user has selected something from the dropdown, I want it to show the Name of the product beside it. I dont even mind if the page has to be refreshed each the user selects something from the drop down menu. I just want it to go to the database on each change and get the name of that product.

Please help!!!!!

Mudley · 18-Jun-2008, 09:24 PM #2

you don't need to go to the database each time as long as the product name is getting pulled in the same query as the product code

use some javascript to change the text next to the select box

use an ONCHANGE event in the <select> tag

do you want to learn how to do this?
or do you just want it done?

Mr.LLB · 19-Jun-2008, 10:05 AM #3

Thanks for your reply,

I would really appreciate if you could only do this for ONE product. This way i get a sample and I get to learn by my self.

Use the following product details:

Product Code: C440
Product Name: Product A.

Product Code: C441
Product Name: Product B

Please use these products to show me an example where by changing the value in the select automatically shows me the product name beside it.

Thanks,

Mr.LLB · 19-Jun-2008, 10:07 AM #4

And lets say that the product code and name are in a database.

Mr.LLB · 19-Jun-2008, 01:45 PM #5

So when we are getting the product details in javascript......they must be called from a database which has someting like 60 products.

help!!!!

TheRobatron · 04:05 PM

This isn't too hard to do - what I would do is read all the product names and codes in one query when the page is loaded. For this example I will assume that they are in arrays called $names and $codes.

The most efficient way to do this is with a for loop (or a foreach loop, but with two separate arrays, I find it easier to keep track of the key with a for loop), populating the dropdown list and javascript as it goes along.

HTML Code:

<head> <script type='text/javascript'>
function textChange(text) {
document.getElementById('nameText').innerHTML = text;
}
</script> </head> <body> <select> <!-- PHP code goes here --> </select> <p id='nameText'></p> </body>

PHP Code:

  for ($x = 0; $x <= sizeof($names); $x = $x + 1) {
echo "<option onSelect=\"textChange('$names[x$]')\">$codes[$x]</option>\n";
}

Mr.LLB · 19-Jun-2008, 04:04 PM #7

Can't i set it up in such a way that it creates the $ name and $code array from the database. Because I will be changing database entries and I dont want to have to manually edit the array????

TheRobatron · 19-Jun-2008, 04:06 PM #8

Yeah, are you using MySQL?

Mr.LLB · 19-Jun-2008, 04:08 PM #9

Mr.LLB · 19-Jun-2008, 04:08 PM **#10**

phpmyadmin as well to manage and create tables, rows, etc

Mr.LLB · 19-Jun-2008, 04:18 PM **#11**

Can i do the following?

$result = mysql_query('SELECT * FROM table') or exit(mysql_error());
while ($row = mysql_fetch_assoc($result)) {$array[$row['P_Codes']] = $row['P_Name'];}

print_r($array);

Mr.LLB · 19-Jun-2008, 04:19 PM **#12**

this should create an array with the product code as the key and the product name as the value.....right?

TheRobatron · 19-Jun-2008, 04:28 PM **#13**

I would do it slightly differently:

PHP Code:

  $result = mysql_query('SELECT * FROM table') or exit(mysql_error());
$x = 0;
while ($row = mysql_fetch_assoc($result)) {
$codes[$x] = $row['P_Codes'];
$names[$x] = $row['P_Name'];
$x = $x + 1;}

Mr.LLB · 19-Jun-2008, 04:35 PM **#14**

Ok this is cool too.....and the javascript stuff you provided earlier will work with this PHP array......right?

TheRobatron · 19-Jun-2008, 04:36 PM **#15**

It should do

If it doesn't, post the code that you're using.

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