MySQL – Dog Adoption Database Design Best Practices

Designing for maximum performance for an anticipated large user-base, I would suggest ensuring bookmarks and tags are not duplicated unnecessarily. Using my suggested schema below, http://google.com/ would become a single bookmark row. Users who want a bookmark to Google would have a row in UserBookmarksTags referencing the Bookmark row and the associated tag row.

I would tend to abstract the UserID away from the Bookmarks table, and use a cross-reference table to determine which users have bookmarks, and what they've tagged each bookmark with.

Using SQL Server, I would do it like this:

USE tempdb;
GO
CREATE TABLE Users
(
    UserID INT NOT NULL 
        CONSTRAINT PK_Users
        PRIMARY KEY CLUSTERED
        IDENTITY(1,1)
    , UserName varchar(255)
);
CREATE TABLE Bookmarks
(
    BookmarkID INT NOT NULL
        CONSTRAINT PK_Bookmarks
        PRIMARY KEY CLUSTERED
        IDENTITY(1,1)
    , BookmarkName varchar(255)
    , BookmarkURL varchar(255)
);
CREATE TABLE Tags
(
    TagID INT NOT NULL
        CONSTRAINT PK_Tags
        PRIMARY KEY CLUSTERED
        IDENTITY(1,1)
    , TagName varchar(255)
);

If every entry in your system requires both a bookmark and a tag, you could use the following table for maximum performance:

CREATE TABLE UsersBookmarksTags
(
    UserID INT NOT NULL
        CONSTRAINT FK_UsersBookmarksTags_UserID
        FOREIGN KEY REFERENCES Users(UserID)
    , TagID INT NOT NULL
        CONSTRAINT FK_UsersBookmarksTags_TagID
        FOREIGN KEY REFERENCES Tags(TagID)
    , BookmarkID INT NOT NULL
        CONSTRAINT FK_UsersBooksmarksTags_BookmarkID
        FOREIGN KEY REFERENCES Bookmarks(BookmarkID)
    , CONSTRAINT PK_UsersBookmarksTags
        PRIMARY KEY CLUSTERED 
        (UserID, TagID, BookmarkID)
);

The above version of the table UsersBookmarksTags allows you to quickly return a list of bookmarks for a specific user that matches any given tag, or all bookmarks for that user.

Alternately, you could create the cross reference table that allows null entries for both Tags and Bookmarks, as in:

CREATE TABLE UsersBookmarksTags
(
    UsersTagsBookmarksID INT NOT NULL
        CONSTRAINT PK_UsersBookmarksTags
        PRIMARY KEY CLUSTERED
        IDENTITY(1,1)
    , UserID INT NOT NULL
        CONSTRAINT FK_UserBookmarks_UserID
        FOREIGN KEY REFERENCES Users(UserID)
    , TagID INT NULL
        CONSTRAINT FK_UserBookmarks_TagID
        FOREIGN KEY REFERENCES Tags(TagID)
    , BookmarkID INT NULL
        CONSTRAINT FK_UserBooksmarks_BookmarkID
        FOREIGN KEY REFERENCES Bookmarks(BookmarkID)
);

This allows rows that have a bookmark without a tag, and also allows tags to be associated to a user without a bookmark.

If there is a requirement to know all the tags a user has added, without those tags necessarily having an associated bookmark, I would consider adding another table for that purpose, as:

CREATE TABLE UserTags
(
    UserID INT NOT NULL
        CONSTRAINT FK_UserTags_UserID
        FOREIGN KEY REFERENCES Users(UserID)
    , TagID INT NOT NULL
        CONSTRAINT FK_UserTags_TagID
        FOREIGN KEY REFERENCES Tags(TagID)
    , CONSTRAINT PK_UserTags
        PRIMARY KEY CLUSTERED 
        (UserID, TagID)
);

To answer your questions:

1. Year number is part of a (three column) composite primary key. There will be one record per employee + year + type of leave (vacation/sick/...) Year number probably means the calendar year stored as an integer.

2. der_leave_cumulative_days_taken means the total of leave actually taken, which would be derived by summing the days between Employee_Leave_Taken.leave_from_date and Employee_Leave_Taken.leave_to_date where those dates fall within the year given by Employee_Annual_Leave.year_number for the same employee_id and leave_type_code.

3. der_leave_days_remaining is next to pointless because it would be Employee_Annual_Leave.leave_days_allowed less der_leave_cumulative_days_taken. A real scenario would be that an employee is allowed three weeks vacation (15 days) and has taken two weeks so far (der_leave_cumulative_days_taken=10 days) and so they have 5 days remaining. The point of keeping these precalculated in this way is that you can refer easily to these values in a query or report. Notice the date_updated field. This implies that there is a regular (perhaps scheduled) batch job that calculates these derived values periodically, or perhaps they are recalculated when changes are made to either leave allowed or leave taken.