Data Modeling, Breaking & Fixing First Normal Form (1NF)

It has been my observation that a lot of us lack the basic
skills required to model the simplest of database objects. This article will
explore some of the common mistakes and fixes to the breaking of 1NF.

1NF Rule

First normal form has two very distinct rules:

1. There
   must not be duplicate columns within a row of a table.

2. There
   must not be more than one value for any column in a table.

In my years of data design there is one over-glaring mistake
that a lot of "newbie" data modelers and database administrators make
that cause enormous amounts of problems for themselves and others in the
development chain. It is their inability to conform to the very basics of data
modeling principles. Of those principles, adhering to 1NF is probably the most broken
rule that I have seen. When the model does not follow 1NF, queries are hard to develop
and producing usable data from the model is even harder. Typically, the
developer will have to put some unusual logic, functions or application code
around or within queries just to make the data manageable.

This article will walk you through these common mistakes and
how you might begin to start working with those mistakes, if you find yourself
in this unfortunate predicament.

Mistake 1, creating another column for a relationship

The first form of breaking 1NF is the creation of another
column in a table, which really just duplicates a relationship to a primary
key. For this example, let’s assume that a distributor wants to keep a
relationship for all of the cities in each of the states that they have a
presence. Our junior modeler quickly creates a CITY table with two columns, one
to hold the state value and the other to hold the city.

This table is modeled quite well but because of the modelers’
inexperience and because he has just learned that there are four other cities
in Colorado that are relevant, he quickly adds four more columns to his CITY
table. The table now looks like the following:

Now you may say to yourself that this isn’t so bad because
we will only ever have 5 cities in the state of Colorado and therefore putting
all of these for quick access in one table is quite efficient. While you are
correct to some extent the real problem comes into play when you wish to
extract the information.

One of the first questions you will get is to list all the
cities in each of the states where the company has a distributor. You could
easily issue the following SQL in Listing 1.
This works just fine but when you try and put the result set into an
application or a developer requests that the information be put into a single
column output so they do not have to traverse the list of cities, the real
problems begin. This request equates to producing output where one city is
represented on one line with the state to which it belongs. To do this there, the
SQL gets a bit interesting and quickly cumbersome to work with. Listing 2 gives just such an example where we
must union together a select statement for each iteration of the cities
contained in the row. The reason for the NOT NULL is because we do not want to
return an empty row where there is not a city for the individual iteration. In
addition, for small tables this is not quite a big deal, but this query will
produce multiple table scans and if there are more than a few rows, you will
have a performance problem.

Just as you start to pat yourself on the back, the next question
you will be asked is if you can provide a query to determine if the company has
a distributor in a particular city. Listing 3
gives you a solution to this question. Remember we must build on the prior
example since typically, this will be coming from a developer and they would
like the output in a single column output. This query can quickly be tailored
to answer the question of in which cities in a particular state do we have
distributors, or how many cities in a particular state do we have a
distributor. This last question is answered in Listing
4.

Listing 1.

Simple method to display cities for states

SQL> Select state, city, city2, city3, city4, city5 from CITY;

STATE      CITY       CITY2      CITY3      CITY4      CITY5

———- ———- ———- ———- ———- ———-

TEXAS      DALLAS     HOUSTON    PARIS

COLORADO   DENVER     BOULDER    ASPEN      MONTROSE   PUEBLO

MICHIGAN   DETROIT    LANSING

Listing 2.

Display cities for states in a true single column output

SQL> select state,city  city from CITY where city  is not null union

     select state,city2 city from CITY where city2 is not null union

     select state,city3 city from CITY where city3 is not null union

     select state,city4 city from CITY where city4 is not null union

     select state,city5 city from CITY where city5 is not null

      order by state;

STATE      CITY

———- ———-

COLORADO   ASPEN

COLORADO   BOULDER

COLORADO   DENVER

COLORADO   MONTROSE

COLORADO   PUEBLO

MICHIGAN   DETROIT

MICHIGAN   LANSING

TEXAS      DALLAS

TEXAS      HOUSTON

TEXAS      PARIS

Listing 3.
Do we have a
distributor in Denver

SQL> select state,city from (

     select state,city  city from CITY where city is not null union

     select state,city2 city from CITY where city2 is not null union

     select state,city3 city from CITY where city3 is not null union

     select state,city4 city from CITY where city4 is not null union

     select state,city5 city from CITY where city5 is not null)

      where city = ‘DENVER’;

STATE      CITY

———- ———-

COLORADO   DENVER

Listing 4.

How many
distributors do we have in each state

SQL> select state,count(*) Distributor_Count from (

     select state,city  city from CITY where city is not null union

     select state,city2 city from CITY where city2 is not null union

     select state,city3 city from CITY where city3 is not null union

     select state,city4 city from CITY where city4 is not null union

     select state,city5 city from CITY where city5 is not null)

      group by state;

STATE      DISTRIBUTOR_COUNT

———- —————–

COLORADO                   5

MICHIGAN                   2

TEXAS                      3