Banner and the SQL Select Statement: Part Three (Joins)

Banner and the SQL Select Statement: Part Three (Joins)

Mark HollidayDepartment of Mathematics and

Computer ScienceWestern Carolina University

4 November 2005 and 11 November 2005(updated: 11 November 2005)

Outline The Goal The Concepts

A First Example Single Table Selects Joins Multiple Connected Select Statements

A First Example

Outline The Relational Model: Single Table Lab 1: TOAD, Schema Browser Some Structured Query Language (SQL)

Basics Lab 2: TOAD, SQL Editor

Single Table Selects

Outline WHERE clause: single condition, multiple

conditions Lab 3: Order By; Aggregate Functions Lab 4: Group By; Having Lab 5:

Joins

Outline Why Multiple Tables? Inner Joins Lab 6: Outer joins Lab 7:

Why Multiple Tables? (franz)Why Multiple Tables? (franz)

One Table database: keeps track of all purchases at our store (known as the ‘flat file’)

Every time a new row is inserted into the table, all columns will be updated. This results in unnecessary "redundant data". For example, every time Wolfgang Schultz purchases something, the following rows will be inserted into the table:

Why Multiple Tables? Why Multiple Tables? (franz)(franz)

What happens if we ONE DAY find out Wolfgang Schultz’s last name is really spelled “S-h-o-o-l-t-z-e” instead of “S-c-h-u-l-t-z”? (franz)He has purchased LOTS of stuff from us.

And…. He has been making purchases at our store for over five years. In fact, he is our very BEST customer. His purchases provide for most of our gross sales.

But now… He says that he will take his business elsewhere if we don’t get this problem corrected!

And YOU have to get it corrected NOW!!!

Why Multiple Tables?Why Multiple Tables?

Hold that thought !Hold that thought !

This helps to exemplify WHY relational database table structures are so nice ...


Original ‘flat file’ table:


For this example, an ideal database would have two tables:

One for keeping track of customer information (customer_info)

And the other to keep track of what they purchase (purchases).

customer_info table:

purchases table:


Our new tables: Our new tables: (franz)(franz)

The ‘id’ numberis the commonelement thatties the tablestogether.

After redesigning our data structure into two tables,

whenever a change for the name or address needs to be made to our repeating customer,

we update the “customer_info” table only.

Only the second table, “purchases”, needs to be updated when purchases are made.


We've just eliminated the space-taking, time-consuming, useless redundant data of our flat file.

That is, we've just NORMALIZED this database!

This is the basis for relational database structure.


Unfortunately, having multiple tables introduces a problem.

Problem: Information you need for a query is often in more than one table.

Example: Find the zip code of the person whose first name is “Ab” and last name is “Mazlan”. need the spraddr table for the zip code and need the spriden table for the name

Joins: IntroductionJoins: Introduction

Solution: Need a way to temporarily (just in the query) connect a row in one table with the row (or rows) in another table that “match” it.

Refined Solution: Use some type of JOIN operation


Problem: How do we find the “matching” rows in the other table?

In other words, how do we do a join?

Solution: 1. Conceptually, match each row in the left

table with every row in the right table.• take the Cartesian Product of the two tables

2. Only keep the row pairs that “match”.


Joins: Introduction (Step One of Join)Joins: Introduction (Step One of Join)

pidm last_name

1 “smith”2 “jones”

pidm zip1 201113 20311

pidm1 Last_name pidm2 zip1 “smith” 1 201111 “smith” 3 203112 “jones” 1 201112 “jones” 3 20311

Cartesian Product

--

Joins: Introduction (Step Two of Join)Joins: Introduction (Step Two of Join)

pidm1 Last_name pidm2 zip1 “smith” 1 201111 “smith” 3 203112 “jones” 1 201112 “jones” 3 20311

only keep rows that have pidm1 = pidm2pidm1 Last_name pidm2 zip1 “smith” 1 20111

Questions:1) When does a row from the left table

“match” a row from the right table?2) What to do with a row in one table

that does not match a row in the other table?


Question One: When does a row from the left table “match” a row from the right table?

Answer: The two rows match if the condition expression is TRUE.

The expression is a sequence of conditions connected by AND.

Each condition is a comparison of a left table column and a right table column

The comparison operator is equality.


Example condition expression just one condition

spriden.spriden_pidm = spraddr.spraddr_pidm


Good relational database design

=> the only common column in two tables is the key for each table

=> use the key of each table to get from a left table row to the matching right table row

• they have the same key value


Question Two: What to do with a row in one table that does not match a row in the other table?

ignore them: inner join don’t ignore them: outer join This section covers inner joins


An example using reserved words to indicate a join:

SELECT spriden_last_name, spriden_first_name, spraddr_city, spraddr_stat_code, spraddr_zip

FROM spriden INNER JOIN spraddrON spriden_pidm = spraddr_pidm

Implicitly, 1) take cartesian product of spriden and spraddr2) only keep the rows that meet the condition

INNER Join (franz)

Syntax for a join using reserved words

SELECT "list-of-columns“FROM table1 [“type” JOIN] table2ON “field matching”WHERE "search-condition(s)"

INNER Join (franz)

Table ‘spriden’:

Table ‘spraddr’:

INNER Join (franz)

SELECT spriden_last_name, spriden_first_name, spraddr_city, spraddr_stat_code, spraddr_zipFROM spriden INNER JOIN spraddrON spriden_pidm = spraddr_pidm

INNER Join (franz)

Task: Select from ‘spriden’ for the active record Task: Select from ‘spriden’ for the active record (spriden_change_ind is null). Pull the permanent mailing (spriden_change_ind is null). Pull the permanent mailing address types (‘MA’) in the US (spraddr_natn_code is null).address types (‘MA’) in the US (spraddr_natn_code is null).

This will pull a large amount of data.

Let’s create a smaller subset by picking zip codes on the northern East Coast (zip codes starting with ‘0’).

Let’s also order our output by the last name then first name.

INNER Join (franz)

To accomplish this we would take the previous SQL and make these additions:

SELECT spriden_last_name, spriden_first_name, spraddr_city, spraddr_stat_code,spraddr_zip

FROM spriden INNER JOIN spraddrON spriden_pidm = spraddr_pidmWHERE spriden_change_ind is nulland spraddr_atyp_code = 'MA'and spraddr_zip between '000%' and '1%'and spraddr_natn_code is nullORDER BY spriden_last_name, spriden_first_name

INNER Join (franz)

SELECT substr(spriden_last_name,1,12) || ', ' || substr(spriden_first_name,1,12), rtrim(spraddr_city) || ', ' || spraddr_stat_code || ' ' || spraddr_zip

FROM spriden INNER JOIN spraddrON spriden_pidm = spraddr_pidmWHERE spriden_change_ind is nulland spraddr_atyp_code = 'MA'and spraddr_zip between '000%' and '1%'and spraddr_natn_code is nullORDER BY spriden_last_name, spriden_first_name

INNER Join (franz)

Part of the Resultant Subset Would Be:

INNER Join (franz)

If we had left out the check for a null change indicator, we would have received duplicate rows of data. (We would get the Gary Abbot record with a null indicator, and also the one with an “I” indicator.)

SELECT substr(spriden_last_name,1,12) || ', ' || substr(spriden_first_name,1,12), rtrim(spraddr_city) || ', ' || spraddr_stat_code || ' ' || spraddr_zip

FROM spriden INNER JOIN spraddrON spriden_pidm = spraddr_pidmWHERE spraddr_atyp_code = 'MA'and spraddr_zip between '000%' and '1%'and spraddr_natn_code is null……spriden_change_ind is nullspriden_change_ind is nullORDER BY spriden_last_name, spriden_first_name

INNER Join (franz)

INNER Join (franz)

SELF JoinSELF JoinThe left table and the right table in an

inner join might be the same table.

This is called a SELF Join.

When would this be useful?

SELF Join (franz)SELF Join (franz)‘spiffy’ table:

Syntax:SELECT e.first_name EMPLOYEE_FIRST, e.last_name EMPLOYEE_LAST, m.first_name MANAGER_FIRST, m.last_name MANAGER_LASTFROM spiffy e INNER JOIN spiffy mON e.manager_id = m.employee_id

SELF JoinSELF Join

Previous Inner Join examples: instances of two different tables joining on the

same column in both tables (e.g. pidm)The Self Inner Join example: two instances of the same table joining on

different columns of the same table (e.g. manager_id and employee_id)

SELECT e.first_name EMPLOYEE_FIRST, e.last_name EMPLOYEE_LAST, m.first_name MANAGER_FIRST, m.last_name MANAGER_LASTFROM spiffy e INNER JOIN spiffy mON e.manager_id = m.employee_id

SELF Join (franz)SELF Join (franz)Result:

Laboratory Six Objectives:

Develop competence with inner joins

Steps: First Query

Laboratory Six Problem: Find the area code and phone

number for every one whose last name is “Holliday.”

Hint: Use the spriden and sprtele tables.

Solution One:SELECT sprtele_phone_area, sprtele_phone_number

FROM spriden INNER JOIN sprteleON spriden_pidm = sprtele_pidmWHERE spriden_last_name = ‘Holliday’

Solution Two:SELECT sprtele_phone_area, sprtele_phone_number FROM spriden, sprteleWHERE spriden_pidm = sprtele_pidm and

spriden_last_name = ‘Holliday’

Laboratory Six

OUTER Join (franz)OUTER Join (franz)

Notice that ‘spriden’ has records for Bagnall, Baker and Barksdale.

‘Spbpers’ has records for Bagnall and Barksdale, but not Baker.


If we join the tables by an inner join on pidm, we would only receive rows for Bagnall and Barksdale.

To pick up all the records in ‘spriden’, and join it with the data in ‘spbpers’ we would use an outer join.


OUTER Joins can be split into three types:OUTER Joins can be split into three types: LEFT outer join RIGHT outer join FULL outer join

“Left” or “Right” designates what your base table is for the join.

“Full” returns all rows in both tables.

SELECT spriden_last_name, spriden_first_name, spbpers_sex, spbpers_birth_date

FROM spriden LEFT OUTER JOIN spbpersON spriden_pidm = spbpers_pidmWHERE spriden_change_ind is nulland spriden_entity_ind = 'P'ORDER BY spriden_last_name, spriden_first_name

LEFT OUTER Join (franz)LEFT OUTER Join (franz)

LEFT OUTER Join (franz)LEFT OUTER Join (franz)

Result:


FROM spriden RIGHT OUTER JOIN spbpersON spriden_pidm = spbpers_pidmWHERE spriden_change_ind is nulland spriden_entity_ind = 'P'ORDER BY spriden_last_name, spriden_first_name

RIGHT OUTER Join (franz)RIGHT OUTER Join (franz)

Result:

RIGHT OUTER Join (franz)RIGHT OUTER Join (franz)


FROM spriden FULL OUTER JOIN spbpersON spriden_pidm = spbpers_pidmWHERE spriden_change_ind is nulland spriden_entity_ind = 'P'ORDER BY spriden_last_name, spriden_first_name

FULL OUTER Join (franz)FULL OUTER Join (franz)

Result: PIDM added to indicate which rows were selected.

FULL OUTER Join (franz)FULL OUTER Join (franz)

Laboratory Seven Objectives:

Develop competence with outer joins

Steps: Outer Join Query

JOIN Query (franz)JOIN Query (franz)Write a JOIN. Use either an “INNER”, “OUTER” or “FULL”.

Use tables SPRIDEN and SPRADDR.

Include both PIDMs, last name, first name, first address line, address type.

Join on PIDMSelect the Active name recordSelect entity indicator for a PersonSelect PIDMs less than 70,000Select last names that start with “Bea”

Order the result set by SPRIDEN PIDM, then SPRADDR PIDM

Answer (franz)Answer (franz)SELECT spriden_pidm, spraddr_pidm,

spriden_last_name, spriden_first_name, spraddr_atyp_code, spraddr_street_line1

FROM spriden LEFT OUTER JOIN spraddrON spriden_pidm = spraddr_pidmWHERE spriden_change_ind is nullAND spriden_entity_ind = 'P'AND spriden_pidm < 70000AND spriden_last_name like 'Bea%'ORDER BY spriden_pidm, spraddr_pidm

Multiple Connected Select Statements

Outline Union, intersection, minus Lab 8: Subqueries

Use directly: FROM clause Use as a set: new operators Use as a single value: aggregate functions

Lab 9: A Query Development Methodology

Documents

Banner and the SQL Select Statement: Part Three (Joins)