Learning SQL Second Edition phần 10 pot

mysql> SELECT fname, lname... mysql> SELECT EXTRACTMONTH FROM CURRENT_DATE; 1 row in set 0.02 sec Your result will most likely be different, unless it happens to be May when you try this

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18 rows in set (0.03 sec)

The correct values for <1> and <2> are:

mysql> SELECT a.account_id, c.fed_id, p.name

-> FROM account a INNER JOIN customer c

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| 18 | 999-99-9999 | money market account |

mysql> SELECT e.emp_id, e.fname, e.lname

-> FROM employee e INNER JOIN employee mgr

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-> SELECT fname, lname

Sort the results from Exercise 6-2 by the lname column.

mysql> SELECT fname, lname

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Write a query that returns the absolute value and sign ( −1 , 0 , or 1 ) of the number −25.

76823 Also return the number rounded to the nearest hundredth.

mysql> SELECT ABS(-25.76823), SIGN(-25.76823), ROUND(-25.76823, 2);

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Write a query to return just the month portion of the current date.

mysql> SELECT EXTRACT(MONTH FROM CURRENT_DATE());

1 row in set (0.02 sec)

(Your result will most likely be different, unless it happens to be May when you try this exercise.)

Chapter 8

8-1

Construct a query that counts the number of rows in the account table.

mysql> SELECT COUNT(*)

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Note that MySQL would not accept ORDER BY SUM(avail_balance) DESC, , so I was forced to indicate the sort column by position.

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Chapter 9

9-1

Construct a query against the account table that uses a filter condition with a related subquery against the product table to find all loan accounts ( product.product_type_cd = 'LOAN' ) Retrieve the account ID, product code, customer

noncor-ID, and available balance.

mysql> SELECT account_id, product_cd, cust_id, avail_balance

-> WHERE p.product_cd = a.product_cd

-> AND p.product_type_cd = 'LOAN');

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UNION ALL

SELECT 'mentor' name, '2000-01-01' start_dt, '2001-12-31' end_dt

Give the subquery the alias levels , and include the employee’s ID, first name, last name, and experience level ( levels.name ) (Hint: build a join condition using an inequality condition to determine into which level the employee.start_date column falls.)

mysql> SELECT e.emp_id, e.fname, e.lname, levels.name

-> FROM employee e INNER JOIN

-> (SELECT 'trainee' name, '2004-01-01' start_dt, '2005-12-31' end_dt

mysql> SELECT e.emp_id, e.fname, e.lname,

-> (SELECT d.name FROM department d

-> WHERE d.dept_id = e.dept_id) dept_name,

-> (SELECT b.name FROM branch b

-> WHERE b.branch_id = e.assigned_branch_id) branch_name

-> FROM employee e;

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mysql> SELECT p.product_cd, a.account_id, a.cust_id, a.avail_balance

-> FROM product p LEFT OUTER JOIN account a

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mysql> SELECT p.product_cd, a.account_id, a.cust_id, a.avail_balance

-> FROM account a RIGHT OUTER JOIN product p

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Outer-join the account table to both the individual and business tables (via the account.cust_id column) such that the result set contains one row per account Col-umns to include are account.account_id , account.product_cd , individual.fname , individual.lname , and business.name

mysql> SELECT a.account_id, a.product_cd,

-> i.fname, i.lname, b.name

-> FROM account a LEFT OUTER JOIN business b

-> ON a.cust_id = b.cust_id

-> LEFT OUTER JOIN individual i

-> ON a.cust_id = i.cust_id;

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10-4 (Extra Credit)

Devise a query that will generate the set {1, 2, 3, , 99, 100} (Hint: use a cross join with at least two from clause subqueries.)

SELECT ones.x + tens.x + 1

FROM

(SELECT 0 x UNION ALL

SELECT 1 x UNION ALL

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SELECT 9 x) ones

CROSS JOIN

(SELECT 0 x UNION ALL

SELECT emp_id,

CASE title

WHEN 'President' THEN 'Management'

WHEN 'Vice President' THEN 'Management'

WHEN 'Treasurer' THEN 'Management'

WHEN 'Loan Manager' THEN 'Management'

WHEN 'Operations Manager' THEN 'Operations'

WHEN 'Head Teller' THEN 'Operations'

WHEN 'Teller' THEN 'Operations'

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-> SUM(CASE WHEN open_branch_id = 1 THEN 1 ELSE 0 END) branch_1,

-> SUM(CASE WHEN open_branch_id = 4 THEN 1 ELSE 0 END) branch_4

START TRANSACTION;

SELECT i.cust_id,

(SELECT a.account_id FROM account a

WHERE a.cust_id = i.cust_id

AND a.product_cd = 'MM') mm_id,

(SELECT a.account_id FROM account a

WHERE a.cust_id = i.cust_id

AND a.product_cd = 'chk') chk_id

INTO @cst_id, @mm_id, @chk_id

FROM individual i

WHERE i.fname = 'Frank' AND i.lname = 'Tucker';

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INSERT INTO transaction (txn_id, txn_date, account_id,

txn_type_cd, amount)

VALUES (NULL, now(), @mm_id, 'CDT', 50);

INSERT INTO transaction (txn_id, txn_date, account_id,

ALTER TABLE account

ADD CONSTRAINT account_unq1 UNIQUE (cust_id, product_cd);

WHERE txn_date > cast('2008-12-31 23:59:59' as datetime);

SELECT txn_date, account_id, txn_type_cd, amount

FROM transaction

WHERE txn_date > cast('2008-12-31 23:59:59' as datetime)

AND amount < 1000;

CREATE INDEX txn_idx01

ON transaction (txn_date, amount);

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| NULL | Michael Smith |

| Michael Smith | Susan Barker |

| Michael Smith | Robert Tyler |

| Robert Tyler | Susan Hawthorne |

| Susan Hawthorne | John Gooding |

| Susan Hawthorne | Helen Fleming |

| Helen Fleming | Chris Tucker |

| Helen Fleming | Sarah Parker |

| Helen Fleming | Jane Grossman |

| Susan Hawthorne | Paula Roberts |

| Paula Roberts | Thomas Ziegler |

| Paula Roberts | Samantha Jameson |

| Susan Hawthorne | John Blake |

| John Blake | Cindy Mason |

| John Blake | Frank Portman |

| Susan Hawthorne | Theresa Markham |

| Theresa Markham | Beth Fowler |

| Theresa Markham | Rick Tulman |

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mysql> CREATE VIEW supervisor_vw

Query OK, 0 rows affected (0.12 sec)

mysql> SELECT * FROM supervisor_vw;

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| supervisor_name | employee_name |

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| NULL | Michael Smith |

| Michael Smith | Susan Barker |

| Michael Smith | Robert Tyler |

| Robert Tyler | Susan Hawthorne |

| Susan Hawthorne | John Gooding |

| Susan Hawthorne | Helen Fleming |

| Helen Fleming | Chris Tucker |

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| Helen Fleming | Sarah Parker |

| Helen Fleming | Jane Grossman |

| Susan Hawthorne | Paula Roberts |

| Paula Roberts | Thomas Ziegler |

| Paula Roberts | Samantha Jameson |

| Susan Hawthorne | John Blake |

| John Blake | Cindy Mason |

| John Blake | Frank Portman |

| Susan Hawthorne | Theresa Markham |

| Theresa Markham | Beth Fowler |

| Theresa Markham | Rick Tulman |

mysql> CREATE VIEW branch_summary_vw

-> SELECT b.name, b.city, sum(a.avail_balance)

-> FROM branch b INNER JOIN account a

-> ON b.branch_id = a.open_branch_id

-> GROUP BY b.name, b.city;

Query OK, 0 rows affected (0.00 sec)

mysql> SELECT * FROM branch_summary_vw;

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| branch_name | branch_city | total_balance |

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| Headquarters | Waltham | 27882.57 |

| Quincy Branch | Quincy | 53270.25 |

| So NH Branch | Salem | 68240.32 |

| Woburn Branch | Woburn | 21361.32 |

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Chapter 15

15-1

Write a query that lists all the indexes in the bank schema Include the table names.

mysql> SELECT DISTINCT table_name, index_name

-> FROM information_schema.statistics

-> WHERE table_schema = 'bank';

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| table_name | index_name |

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| account | PRIMARY |

| account | account_unq1 |

| account | fk_product_cd |

| account | fk_a_branch_id |

| account | fk_a_emp_id |

| account | acc_bal_idx |

| branch | PRIMARY |

| business | PRIMARY |

| customer | PRIMARY |

| department | PRIMARY |

| department | dept_name_idx |

| employee | PRIMARY |

| employee | fk_dept_id |

| employee | fk_e_branch_id |

| employee | fk_e_emp_id |

| individual | PRIMARY |

| officer | PRIMARY |

| officer | fk_o_cust_id |

| product | PRIMARY |

| transaction | PRIMARY |

| transaction | fk_t_account_id |

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15-2

Write a query that generates output that can be used to create all of the indexes on the bank.employee table Output should be of the form:

"ALTER TABLE <table_name> ADD INDEX <index_name> (<column_list>)"

mysql> SELECT concat(

-> CASE

-> WHEN st.seq_in_index = 1 THEN

-> concat('ALTER TABLE ', st.table_name, ' ADD',

-> CASE

-> WHEN st.non_unique = 0 THEN ' UNIQUE '

-> ELSE ' '

-> END,

-> 'INDEX ',

-> st.index_name, ' (', st.column_name)

-> ELSE concat(' ', st.column_name)

-> END,

-> CASE

-> WHEN st.seq_in_index =

-> (SELECT max(st2.seq_in_index)

-> FROM information_schema.statistics st2

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-> WHERE st2.table_schema = st.table_schema

-> AND st2.table_name = st.table_name

-> AND st2.index_name = st.index_name)

-> THEN ');'

-> ELSE ''

-> END

-> ) index_creation_statement

-> FROM information_schema.statistics st

-> WHERE st.table_schema = 'bank'

-> AND st.table_name = 'employee'

-> ORDER BY st.index_name, st.seq_in_index;

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| index_creation_statement |

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| ALTER TABLE employee ADD INDEX fk_dept_id (dept_id); |

| ALTER TABLE employee ADD INDEX fk_e_branch_id (assigned_branch_id); | | ALTER TABLE employee ADD INDEX fk_e_emp_id (superior_emp_id); |

| ALTER TABLE employee ADD UNIQUE INDEX PRIMARY (emp_id); |

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using with filter conditions, 64

< (less than) operator

scalar subqueries and, 159

using with all operator, 163

<= (less than or equal to) operator, 159

<> (not equal to) operator

in inequality conditions, 67

= (equals sign)

= null, filtering for null values, 77

equal to operator

using with any operator, 165

in equality conditions, 67

> (greater than) operator

>= (greater than or equal to) operator, 159

\ (backslash), escaping special characters in

strings, 116 _ (underscore), wildcard character in partial

string matches, 74

A

abs( ) function, 130 aggregate functions, 144, 145–150 count( ) function, 147

exercises with, 156 handling null values, 149

in having clause, 155 implicit versus explicit groups, 146 listing of common functions, 145 using expressions as arguments, 149 where clause and, 155

aggregation selective aggregation using case expressions,

209 using views for data aggregation, 249 all operator, 163

<> all comparisons, null values and, 163 alter table statements

adding or removing constraints, 239 adding or removing indexes, 229 changing storage engine, 224 modifying definition of existing table, 31 and operator

condition evaluation with, 63 three-condition evaluation using and, or,

64 three-condition evaluation using and, or,

and not, 65 using in select statement where clause, 54 ANSI mode, 115

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Tiêu đề	Learning SQL Second Edition phần 10 pot
Trường học	WoweBook
Chuyên ngành	SQL Database
Thể loại	Sách hướng dẫn

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