Posting info on a page depending on value chosen in selectbox - php

Question

I am working with creating an ad posting page. I have included dependable select boxes as the category selection system. All the values are being pulled from MySQL database. The table is composed of an id (auto increment) bridged by master_id and identified by category name. Every category has a subcategory. So the user is forced to pick a subcategory before being able to submit the info. I am currently looking for ideas of how to post info that has been input by user depending on the category chosen. I have created an individual page for each subcategory. So every time an input is submitted, I want it to display in the page accordingly. For example, if I pick category: Books and subcategory:Textbooks then post it in textbooks.php. Live EXAMPLE
Table Structure
Drop down menus
<form action="" method="post">
<select name="category" id="category" size="7" class="updateCategory" >
<option selected="selected" value="">Select one:</option>
<?php if (!empty($list)) { ?>
<?php foreach($list as $row) { ?>
<option value="<?php echo $row['id']; ?>">
<?php echo $row['name']; ?>
</option>
<?php } ?>
<?php } ?>
</select>
<select name="subc1" id="subc1" size="7" class="updateCategory"
disabled="disabled" hidden="hidden" >
<option value="">----</option>
</select>
<select name="subc2" id="subc2" size="7" class="updateCategory"
disabled="disabled" hidden="hidden">
<option value="">----</option>
</select>
<select name="subc3" id="subc3" size="7" class="updateCategory"
disabled="disabled" hidden="hidden">
<option value="">----</option>
</select>
</br>
Insert your name: <input type="text" name="lname"><br>
<input type="submit" value="Submit">
</form>

Answer 1

This is a little bit of a shot in the dark, as the question was vaguely worded.
I think that, first, you will need to make an update to the database. You will need to create either a new table or a new column. I would personally choose the table as that provides more flexibility, like so: id, category_id, category_page (and you could put other columns in too). Obviously, category_id references back to the main table category.
Then, do a select like this:
SELECT [COLUMNS HERE] FROM category RIGHT JOIN category_page ON category_page.category_id = category.id
You now have the values you need.
Then, in your PHP, return the url to the page (if you are using JS to dynamically show the new page):
<option value="xyz" data-url="url-for-page.php">Category Name</option>