Multiple Variables In A Drop Down List?

Jul 1, 2011

I am creating a quote calculator as a mobile app using the JQuery Mobile plugin in Dreamweaver CS5.5. This supports HTML5 & Javascript.

I have included a drop down list in which the user selects a specific Annual Volume. I have 2 different calculations that need to be done based on the users selection. Different calculations meaning that I need 2 different number values assigned to the option.

For Example:

Calculation #1 (I'm calculating what the Run Quantity is based on what the user selects as the Annual Volume. If user selects an Annual Volume of 150,000 then the calculated result for Run Quantity needs to be 3500)

HTML for this scenario:
<li data-role="fieldcontain">
<p><span class="ui-listview-inset">Annual Volume:</label>
<select name="annualvolume" id="annualvolume">
onChange="Calculate();">

[Code].....

Currently I am simply having the user enter the annual volume twice so I can do the 2 calculations, but this is really clunky and not ideal.

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I have a table that has parts with multiple characteristics. Part Number, Height, Width, Capacity, Price, Etc. I would like to have a drop downs for all of these values, when a user selects one of these values it will filter the results into a list. Ex. user selects a part that has a Height of 6 and width of 10 it will only list those parts. As I said earlier, I have a script that will give me all the parts, however I am not able to make the onchange list parts with multiple values.

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I was wondering how you would list multiple variables after a equal ==?
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Or even list multiple URL's within the variables?
Example
....
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[Code]...

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I am asking jQuery to tell me which drop down option is selected in a drop down list - like this:

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If no id exists.
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In these cases am I expecting a null return; an undefined return, a false return value?

And, based upon the complete set of return possibilities, what would be the best and most comprehensive tests I could apply to cover every base.

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<?php

$link = mysql_connect('myhost', 'myusername', 'mypassword') or die('Could not connect: ' . mysql_error());
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[Code]...

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something like this:

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I'm having a problem passing a variable through a URL because the
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I'm trying to pass data to my mySQL database, but I'm doing that 2 jQuery scripts:User comes on my page where there is a list with some values, e.g 530, 532, 534 etc etc ..User clicks on one of them, this link is using the jQuery to get another php file which get's the data from my DB and list it below the first list, here's the code for this:

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[code].....

So the second list is another step that can't be avoided to get the final result. In other words, i'ts another list where the user has to chose one of the items from the list to see the final result, I've added this code inf the 'e.php' file:

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[code].....

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Code:
function ajaxRefresh(){
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Example:
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I'd like to have multiple option values:
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Code:
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Code:
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[code]....

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