Interfacing Relational Databases to the Web

PostgreSQL uses a different access system than the rest of your system; oddly enough, not even root has access to the database by default. The database system has its own user system and passwords, and postgres is the database administrator account by default. The advantage to the separate access system is that one can create database users who do not have UNIX accounts; this way, the database for your web application can specify access control without creating a potential security hole for your system. To add your web administrator (web) as a database user, use createuser (as root):

# su postgres -c createuser
Enter name of user to add ---> web
Enter user's postgres ID, or RETURN to use UNIX user id: 542 -> 542
Is user "web" allowed to create databases (y/n) y
Is user "web" allowed to add users? (y/n) y
createuser: web was successfully added

Then, as web (or whatever account you used), you'll be able to create a database with createdb foo and then try some queries on foo using psql foo.

You'll also need to set up PostgreSQL to accept incoming TCP/IP connections so your PHP3 pages can access it. Fortunately, System V init makes this easy. Simply open the file /etc/rc.d/init.d/postgresql and change the line

su postgres -l -c \
 'usr/bin/postmaster -S -D/var/lib/pgsql'

so that it reads

su postgres -l -c \
 '/usr/bin/postmaster -S -D/var/lib/pgsql -i'

Just about every relational database in the world uses SQL (or some extended version of SQL) as its query language. SQL allows you to define tables, select records based on given criteria, update values in one or many records and delete records. This is just a brief introduction to SQL; for more complete references, see Resources.

To create a table, one uses the CREATE TABLE statement. Its syntax is as follows:

CREATE TABLE tablename (field-1 type-1, ..., field-n type-n)

In psql, you will need to end each statement with a semicolon. These semicolons are not part of the SQL language, but rather for the benefit of psql's lexer.

You may also declare fields as NOT NULL, UNIQUE or PRIMARY KEY, or specify a value as DEFAULT to a field. PostgreSQL will create an index on primary key fields. Unfortunately, as of version 6.4, PostgreSQL does not support foreign keys, but at least the parser will not choke on the SQL REFERENCES keyword.

Here's an example, akin to the UNIX password file:

CREATE TABLE passwd
   (username    varchar(8) PRIMARY KEY,
   -- PRIMARY KEY implies UNIQUE
    cryptedpass char(13),
    uid         int UNIQUE NOT NULL,
    gid         int NOT NULL,
    gecos       varchar(80),
    -- the GECOS field (real name, office, etc.)
    homedir     varchar(80),
    shell       varchar(50) DEFAULT '/bin/sh');

Note that SQL uses a double-dash to begin comments, which are terminated by a newline.

To insert data into a table, use the INSERT statement:

INSERT INTO tablename (field-1, ..., field-n)
VALUES (value-1, ..., value-n/)

You needn't specify field names if you are inserting values into every field. Here's an example for the table we just created:

INSERT INTO passwd (username, cryptedpass,
   uid, gid, gecos, homedir, shell)
VALUES ('fred', '37MniLTaiPLaL', 42, 500,
   'Fred Mbogo', '/home/fred/', '/bin/sh');

The SQL SELECT statement returns records where values meet a certain criteria. Here are some examples of SELECT in action:

SELECT * FROM passwd;
-- returns all fields of all records
SELECT username FROM passwd;
-- returns all usernames
SELECT * FROM passwd WHERE username = 'fred';
SELECT * FROM passwd ORDER BY username, shell;
SELECT * FROM passwd
     WHERE homedir LIKE '/home%'
     -- % is the SQL wildcard character
     AND   shell = '/bin/sh'
     ORDER BY username;
SELECT homedir, projectname
     FROM passwd, projects
     -- assuming we have a projects
     table
     WHERE
     -- this will return the home directory of
       passwd.username = projects.leader;
       -- each project leader for each project