需要基于主MySql数据库中的数据是否已经存在来填充菜单的可用选项

Question

I've been searching for hours. I doubt this is a unique situation - I just don't know where to look or how to build a proper Query to MySql.

Main data table is for a website CMS. So, there's the PAGE it's on, the SECTION of the page it shows up on, and the CONTENT (text, html, etc)

I then have two tables, one having all the available PAGE names on the site, and one with all the possible SECTION names. (used to populate the menus)

In the CMS data New Entry page, I want the user to be able to select from drop-down menus the PAGE they want first, and (here's the tricky part) then populate the second drop-down menu with ONLY the page sections that don't already have data. (So they can't create a second "Index Page Main Content" record, for example).

I've found plenty of solutions using all manner of things to dynamically fill the second menu - but no clue how to create a query that polls the main db based on the page chosen from the first menu and looks to see what sections types are still unused.

As per request, http://sqlfiddle.com/#!2/736600 for the database setups.

Answer 1

See my SQLFiddle at http://sqlfiddle.com/#!2/1573b/5 .

CREATE TABLE a 
(
 id int auto_increment primary key, 
 page varchar(20), 
 section varchar(20),
 content varchar(30)
);

INSERT INTO a
(page, section,content)
VALUES
('home', 'main', 'hello'),
('home', 'sidebar', 'hello'),
('about', 'main', 'hello'),
('about', 'sidebar', 'hello')

List pages: select DISTINCT(page) from a;

List sections: select DISTINCT(section) from a;

List sections for a specific page: select DISTINCT(section) from a where page = 'home'

You should be able to take the results of the last query, compare it to your list of sections and show only items that do not exist from the SQL query result.