hi I want to be able to calculate the this:I have a date column in mysql, this is the date when a product is added, then the user enters how many days a product must show, this is stored in days column, so I want to show all products that are from date column plus the days chosen in the days column, or all products that don't go pass the days chossen. any clue?? sort of stuck in this one!!lastly can anyone give me any performance tips in php? is there any core rules?thanks!!

If I read that right, there are to ways to do it.

1. Determine the expiration when the item is entered and enter that value into a different field which is checked before the item is displayed.

2. Read the existing mysql values and script it so that
// If days are more than x from insertion date do not display

Frankly, I might be able to write the first one. The second is beyond my current ability.

I think I can help you with this, please contact me if you still need help with this problem.
I have an example of what I think you are asking for.
The example I have to show you is live online currently.

@seo_man, that's nice

@Necaxa, Assuming you have a datetime column called `date_added` and an integer column called `days_to_show` in a table called `products` here's the SQL statement you need.

SELECT *
FROM `products`
WHERE ADDDATE(`date_added`, `days_to_show`) >= CURDATE()

Code seems pretty self explanatory to me, but if you need anything explained, give a shout out.

Use the SQL server to do the calculation

SELECT fieldlist FROM table WHERE CURDATE() BETWEEN datecolumn AND DATE_ADD(datecolumn, INTERVAL daycountcolumn, DAY);

not tested but looks ok

thanks a lot, I'll give it a try and get back to you.here is something abit more complicated....I am using ajax to submit form.in the form you can upload pics, so when I press "upload" the pics are uploaded ,then press "send" and the values of the file/text box are sent to db. in my local pc I use $location ="upload/".basename(textbox) then on my db I get "upload/bla.jpg", now I just tested it on server and it gives this "upload/C:/folder/bla.jpg"? any ideas? ajax is tricky! I tried to put the name of the image in a hidden field in the upload script but is not being posted, so I am stuck! lastly any of you guys know how to speed up a site?