Simulado Prova 1Z0-071: Oracle Database OCA 12c

Question

See the Exhibit and Examine the structure of the CUSTOMERS table: <IMAGEM> Using the CUSTOMERS table, you need to generate a report that shows an increase in the credit limit by 15% for all customers. Customers whose credit limit has not been entered should have the message "Not Available" displayed. Which SQL statement would produce the required result?

Answer 1

SELECT NVL(cust_credit_limit,'Not Available')*.15 "NEW CREDIT" FROM customers;

Answer 2

SELECT NVL(cust_credit_limit*.15,'Not Available') "NEW CREDIT" FROM customers;

Answer 3

SELECT TO_CHAR(NVL(cust_credit_limit*.15,'Not Available')) "NEW CREDIT" FROM customers;

Answer 4

SELECT NVL(TO_CHAR(cust_credit_limit*.15),'Not Available') "NEW CREDIT" FROM customers;

Answer 5

SELECT member_id AS MEMBER_ID, due_date AS DUE_DATE, $2 AS LATE_FEE FROM BOOKS_TRANSACTIONS;

Answer 6

SELECT member_id 'MEMBER_ID', due_date 'DUE_DATE', '$2 AS LATE_FEE' FROM BOOKS_TRANSACTIONS;

Answer 7

SELECT member_id AS "MEMBER_ID", due_date AS "DUE_DATE", '$2' AS " LATE_FEE" FROM BOOKS_TRANSACTIONS;

Answer 8

SELECT member_id AS "MEMBER_ID", due_date AS "DUE_DATE", $2 AS "LATE_FEE" FROM BOOKS_TRANSACTIONS;

Answer 9

You can use column alias in the GROUP BY clause.

Answer 10

Using the WHERE clause after the GROUP BY clause excludes the rows after creating groups.

Answer 11

The GROUP BY clause is mandatory if you are using an aggregate function in the SELECT clause.

Answer 12

Using the WHERE clause before the GROUP BY clause excludes the rows before creating groups.

Answer 13

If the SELECT clause has an aggregate function, then those individual columns without an aggregate function in the SELECT clause should be included in the GROUP BY clause.

Answer 14

Only one LONG column can be used per table.

Answer 15

ATIMESTAMP data type column stores only time values with fractional seconds.

Answer 16

The BLOB data type column is used to store binary data in an operating system file.

Answer 17

The minimum column width that can be specified for a varchar2 data type column is one.

Answer 18

The value for a CHAR data type column is blank-padded to the maximum defined column width.

Answer 19

SELECT d.department_id, c.course_id FROM department_details d RIGHT OUTER JOIN course_details c ON (d.department_id=c.deparment_id);

Answer 20

SELECT d.department_id, c.course_id FROM department_details d LEFT OUTER JOIN course_details c ON (d.department_id=c.deparment_id);

Answer 21

SELECT d.department_id, c.course_id FROM course_details c LEFT OUTER JOIN department_details d ON (c.department_id=d.deparment_id);

Answer 22

SELECT d.department_id, c.course_id FROM department_details d RIGHT OUTER JOIN course_details c ON (c.department_id=d.deparment_id);

Answer 23

The data dictionary is created and maintained by the database administrator.

Answer 24

The data dictionary views can consist of joins of dictionary base tables and user-defined tables.

Answer 25

The usernames of all the users including the database administrators are stored in the data dictionary.

Answer 26

The USER_CONS_COLUMNS view should be queried to find the names of the columns to which a constraint applies.

Answer 27

Both USER_OBJECTS and CAT views provide the same information about all the objects that are owned by the user.

Answer 28

Views with the same name but different prefixes, such as DBA, ALL and USER, use the same base tables from the data dictionary.

Answer 29

COUNT(*) returns the number of rows including duplicate rows and rows containing NULL value in any of the columns

Answer 30

COUNT(cust_id) returns the number of rows including rows with duplicate customer IDs and NULL value in the CUST_ID column

Answer 31

COUNT(DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULL values in the INV_AMT column

Answer 32

A SELECT statement using COUNT function with a DISTINCT keyword cannot have a WHERE clause

Answer 33

The COUNT function can be used only for CHAR, VARCHAR2 and NUMBER data types

Answer 34

It ignores NULL values

Answer 35

The number of columns and data types must be identical for all SELECT statements in the query

Answer 36

The names of columns in all SELECT statements must be identical

Answer 37

Reversing the order of the intersected tables the result

Answer 38

It would be automatically enabled and deferred.

Answer 39

It would be automatically enabled and immediate.

Answer 40

It would remain disabled and has to be enabled manually using the ALTER TABLE command.

Answer 41

It would remain disabled and can be enabled only by dropping the foreign key constraint and recreating it.

Answer 42

When a table is dropped, the corresponding indexes are automatically dropped.

Answer 43

For each DML operation performed, the corresponding indexes are automatically updated.

Answer 44

Indexes should be created on columns that are frequently referenced as part of an expression.

Answer 45

A non-deferrable PRIMARY KEY or UNIQUE KEY constraint in a table automatically creates a unique index.

Answer 46

DELETE order_id FROM orders WHERE order_total < 1000;

Answer 47

DELETE orders WHERE order_total < 1000;

Answer 48

DELETE FROM orders WHERE (SELECT order_id FROM order_items);

Answer 49

DELETE orders o, order_items i WHERE o.order id = i.order id;

Answer 50

A foreign key cannot contain NULL values.

Answer 51

A column with the UNIQUE constraint can contain NULL.

Answer 52

A constraint is enforced only for the INSERT operation on a table.

Answer 53

A constraint can be disabled even if the constraint column contains data.

Answer 54

All the constraints can be defined at the column level as well as the table level.

Answer 55

It recovers only the table structure.

Answer 56

It recovers the table structure, data, and the indexes.

Answer 57

It recovers the table structure and data but not the related indexes.

Answer 58

It is not possible to recover the table structure, data, or the related indexes.

Answer 59

SELECT * FROM MEMBERS WHERE state LIKE '%A_* ;

Answer 60

SELECT * FROM MEMBERS WHERE state LIKE 'A_*;

Answer 61

SELECT * FROM MEMBERS WHERE state LIKE 'A_%';

Answer 62

SELECT * FROM MEMBERS WHERE state LIKE 'A%';

Answer 63

It displays 5 percent of the products with the highest amount sold.

Answer 64

It displays the first 5 percent of the rows from the SALES table.

Answer 65

It displays 5 percent of the products with the lowest amount sold.

Answer 66

It results in an error because the ORDER BY clause should be the last clause.

Answer 67

Outer join

Answer 68

CREATE INDEX

Answer 69

CREATE SESSION

Answer 70

SELECT ANY TABLE

Answer 71

Tablespaces

Answer 72

The product_id value for the first record displayed is 220.

Answer 73

The product_id values for the last two rows displayed are 140 and 126.

Answer 74

The description value for the first two rows displayed is C 2pk-battery.

Answer 75

The description value for the first two rows displayed is AA 2pk-battery.

Answer 76

No row with a product_id of 220is displayed.

Answer 77

The SYSDATE function returns the local machine date and time.

Answer 78

The NVL single-row function can be used on VARCHAR2 columns.

Answer 79

The LENGTH character function returns the number of characters in an expression.

Answer 80

The ROUND number function returns a number rounded to the specified column value.

Answer 81

The TRUNC date function returns a date with the time portion of the day truncated to the specified format unit.

Answer 82

The SUBSTR character function replaces a portion of a string, beginning at a defined character position for a defined length.

Answer 83

Single-row subqueries can only be used in a WHERE clause.

Answer 84

Multiple-row subqueries can be used with the LIKE operator.

Answer 85

Single-row operators can only be used with single-row subqueries.

Answer 86

Single- and multiple-row subqueries can be used with the BETWEEN operator.

Answer 87

Multiple-row subqueries can be used with both single-row and multiple-row operators.

Answer 88

Multiple-row subqueries can be used in a WHERE clause and the INTO portion of an INSERT statement.

Answer 89

All data along with the table structure is deleted.

Answer 90

A pending transaction in the session is committed.

Answer 91

All indexes on the table remain but they are invalidated.

Answer 92

All views and synonyms on the table remain but they are invalidated.

Answer 93

All data in the table is deleted but the table structure remains.

Answer 94

none, because no input required

Answer 95

both the substitution variables ''hiredate' and 'mgr_id'.

Answer 96

only 'hiredate'

Answer 97

only 'mgr_id'

Answer 98

It generates an error because multiple columns (SALARY, COMMISSION) cannot be specified together in an UPDATE statement.

Answer 99

It generates an error because a subquery cannot have a join condition in a UPDATE statement.

Answer 100

It executes successfully and gives the correct result.

Answer 101

It executes successfully but does not give the correct result.

Answer 102

Performance would improve in query 2 only if there are null values in the CUST_CREDIT_LIMIT column.

Answer 103

There would be no change in performance.

Answer 104

Performance would degrade in query 2.

Answer 105

Performance would improve in query 2.

Answer 106

The ERD must have a 1:M relationship between the STUDENTS and PROJECTS entities.

Answer 107

The ERD must have a M:M relationship between the STUDENTS and PROJECTS entities that must be resolved into 1:M relationships.

Answer 108

STUDENT_ID must be the primary key in the STUDENTS entity and foreign key in the PROJECTS entity.

Answer 109

PROJECT_ID must be the primary key in the PROJECTS entity and foreign key in the STUDENTS entity.

Answer 110

An associative table must be created with a composite key of STUDENT_ID and PROJECT_ID, which is the foreign key linked to the STUDENTS and PROJECTS entities.

Answer 111

It would not execute because the same column has been used in both sides of the AND logical operator to form the condition.

Answer 112

It would not execute because the entire WHERE clause condition is not enclosed within the parentheses.

Answer 113

It would execute and the output would display the desired result.

Answer 114

It would execute but the output would return no rows.

Answer 115

The outer query stops evaluating the result set of the inner query when the first value is found.

Answer 116

It is used to test whether the values retrieved by the inner query exist in the result of the outer query.

Answer 117

It is used to test whether the values retrieved by the outer query exist in the result set of the inner query.

Answer 118

The outer query continues evaluating the result set of the inner query until all the values in the result set are processed.

Answer 119

SELECT name, concat (address| | ','| |city| |', ', country) AS full_address, start_date, property_price, property_price*115/100 FROM stores WHERE MONTHS_BETWEEN (start_date, '01-JAN-2000') <=36;

Answer 120

SELECT name, concat (address| | ','| |city| |', ', country) AS full_address, start_date, property_price, property_price*115/100 FROM stores WHERE TO_NUMBER(start_date-TO_DATE('01-JAN-2000','DD-MON-RRRR')) <=36;

Answer 121

SELECT name, address||','||city||','||country AS full_address, start_date, property_price, property_price*115/100 FROM stores WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR')) <=36;

Answer 122

SELECT name, concat (address||','| |city| |', ', country) AS full_address, start_date, property_price, property_price*115/100 FROM stores WHERE MONTHS_BETWEEN (start_date, TO_DATE('01-JAN-2000','DD-MON-RRRR')) <=36;

Answer 123

The execution fails unless the numeral 3 in the ORDER BY clause is replaced by a column name.

Answer 124

Rows are displayed in the order that they are stored in the table only for the three rows with the lowest values in the key column.

Answer 125

Rows are displayed in the order that they are stored in the table only for the first three rows.

Answer 126

Rows are displayed sorted in ascending order of the values in the third column in the table.

Answer 127

When the BOOKS table is dropped, the BOOK_TRANSACTIONS table is dropped.

Answer 128

When the BOOKS table is dropped, all the rows in the BOOK_TRANSACTIONS table are deleted but the table structure is retained.

Answer 129

When a row in the BOOKS table is deleted, the rows in the BOOK_TRANSACTIONS table whose BOOK_ID matches that of the deleted row in the BOOKS table are also deleted.

Answer 130

When a value in the BOOKS.BOOK_ID column is deleted, the corresponding value is updated in the BOOKS_TRANSACTIONS.BOOK_ID column.

Answer 131

SELECT m.last_name "Manager", e.last_name "Employee" FROM employees m JOIN employees e ON m.employee_id = e.manager_id WHERE m.manager_id = 100;

Answer 132

SELECT m.last_name "Manager", e.last_name "Employee" FROM employees m JOIN employees e ON m.employee_id = e.manager_id WHERE e.manager_id = 100;

Answer 133

SELECT m.last_name "Manager", e.last_name "Employee" FROM employees m JOIN employees e ON e.employee_id = m.manager_id WHERE m.manager_id = 100;

Answer 134

SELECT m.last_name "Manager", e.last_name "Employee" FROM employees m JOIN employees e WHERE m.employee_id = e.manager_id and AND e.manager_id = 100;

Answer 135

They can contain a subquery within a subquery.

Answer 136

They can return multiple columns as well as rows.

Answer 137

They cannot contain a subquery within a subquery.

Answer 138

They can return only one column but multiple rows.

Answer 139

They can contain group functions and GROUP BY and HAVING clauses.

Answer 140

They can contain group functions and the GROUP BY clause, but not the HAVING clause.

Answer 141

SELECT e.last_name, m.manager_id FROM employees e RIGHT OUTER JOIN employees m on (e.manager_id = m.employee_id) AND e.employee_id = 123;

Answer 142

SELECT e.last_name, m.manager_id FROM employees e RIGHT OUTER JOIN employees m on (e.employee_id = m.manager_id) WHERE e.employee_id = 123;

Answer 143

. SELECT e.last_name, e.manager_id FROM employees e RIGHT OUTER JOIN employees m on (e.employee_id = m.employee_id) WHERE e.employee_id = 123;

Answer 144

SELECT m.last_name, e.manager_id FROM employees e LEFT OUTER JOIN employees m on (e.manager_id = m.manager_id) WHERE e.employee_id = 123;

Answer 145

second normal form

Answer 146

first normal form

Answer 147

third normal form

Answer 148

fourth normal form

Answer 149

After executing the ALTER TABLE command, you can add a new column called ORDER_DATE to the ORDERS table.

Answer 150

he ORDER_DATE column should be empty for the ALTER TABLE command to execute succsessfully.

Answer 151

ROLLBACK can be used to get back the ORDER_DATE column in the ORDERS table.

Answer 152

The DESCRIBE command would still display the ORDER_DATE column.

Answer 153

Multiple columns or expressions can be compared between the main query and subquery.

Answer 154

Subqueries can contain ORDER BY but not the GROUP BY clause.

Answer 155

Main query and subquery can get data from different tables.

Answer 156

Subqueries can contain GROUP BY and ORDER BY clauses.

Answer 157

Main query and subquery must get data from the same tables.

Answer 158

Only one column or expression can be compared between the main query and subquery.

Answer 159

In a character sort, the values are case-sensitive.

Answer 160

NULL values are not considered at all by the sort operation.

Answer 161

Only those columns that are specified in the SELECT list can be used in the ORDER BY clause.

Answer 162

Numeric values are displayed from the maximum to the minimum value if they have decimal positions.

Answer 163

SELECT last_name, city FROM members WHERE state ='MO' AND state ='MI';

Answer 164

SELECT last_name, city FROM members WHERE state LIKE 'M%';

Answer 165

SELECT last_name, city FROM members WHERE state IN ('MO', 'MI');

Answer 166

SELECT DISTINCT last_name, city FROM members WHERE state ='MO' OR state ='MI';

Answer 167

adding a column constraint when inserting a row into a table

Answer 168

adding a column with a default value when inserting a row into a table

Answer 169

removing all data only from one single column on which a unique constraint is defined

Answer 170

removing all data only from one single column on which a primary key constraint is defined

Answer 171

ALTER TABLE orders MODIFY CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);

Answer 172

ALTER TABLE orders ADD CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);

Answer 173

ALTER TABLE orders MODIFY customer_id CONSTRAINT orders_cust_nn NOT NULL (customer_id);

Answer 174

ALTER TABLE orders ADD customer_id NUMBER(6)CONSTRAINT orders_cust_id_nn NOT NULL;

Answer 175

SELECT inv_no, NVL2(inv_date, 'Pending', 'Incomplete') FROM invoice;

Answer 176

SELECT inv_no, NVL2(inv_amt, inv_date, 'Not Available') FROM invoice;

Answer 177

SELECT inv_no, NVL2(inv_date, sysdate-inv_date, sysdate) FROM invoice;

Answer 178

SELECT inv_no, NVL2(inv_amt, inv_amt*.25, 'Not Available') FROM invoice;

Answer 179

A column can be dropped only if it does not contain any data.

Answer 180

A column can be dropped only if another column exists in the table.

Answer 181

A dropped column can be rolled back.

Answer 182

The column in a composite PRIMARY KEY with the CASCADE option can be dropped.

Answer 183

A parent key column in the table cannot be dropped.

Answer 184

SELECT COUNT (DISTINCT list_price) FROM product_information WHERE list_price is NULL

Answer 185

SELECT COUNT (NVL(list_price, 0)) FROM product_information WHERE list_price is NULL

Answer 186

SELECT COUNT (list_price) FROM product_information WHERE list_price i= NULL

Answer 187

SELECT COUNT (list_price) FROM product_information WHERE list_price is NULL

Answer 188

displaying a date in a nondefault format

Answer 189

finding the number of characters in an expression

Answer 190

substituting a character string in a text expression with a specified string

Answer 191

combining more than two columns or expressions into a single column in the output

Answer 192

PUBLIC should be replaced with specific usernames.

Answer 193

ALL should be replaced with a list of specific privileges.

Answer 194

WITH GRANT OPTION should be added to the statement.

Answer 195

Separate GRANT statements are required for ORDERS and ORDER_ITEMS tables.

Answer 196

The EXECUTION_DATETIME must be of INTERVAL DAY TO SECOND data type.

Answer 197

The EXECUTION_DATETIME must be of TIMESTAMP data type.

Answer 198

The EXECUTION_DATETIME must be of DATE data type.

Answer 199

The COMPONENT_ID must be of ROWID data type.

Answer 200

The COMPONENT_ID must be of VARCHAR2 data type.

Answer 201

The COMPONENT_ID column must be of CHAR data type.

Answer 202

In generates an error because rrrr should be replaced by rr in the format string.

Answer 203

It executes successfully but does not return the correct result.

Answer 204

It executes successfully and returns the correct result.

Answer 205

In generates an error because TO_CHAR should be replaced with TO_DATE.

Answer 206

In generates an error because fm and double quotation marks should not be used in the format string.

Answer 207

You can use column alias in the GROUP BY clause.

Answer 208

Using the WHERE clause after the GROUP BY clause excludes the rows after creating groups.

Answer 209

The GROUP BY clause is mandatory if you are using an aggregate function in the SELECT clause.

Answer 210

Using the WHERE clause before the GROUP BY clause excludes the rows before creating groups.

Answer 211

If the SELECT clause has an aggregate function, then those individual columns without an aggregate function in the SELECT clause should be included in the GROUP BY cause.

Answer 212

Only statement 2 executes successfully and gives the required result.

Answer 213

Only statement 1 executes successfully and gives the required result.

Answer 214

Both statements 1 and 2 execute successfully and give different results.

Answer 215

Both statements 1 and 2 execute successfully and give the same required result.

Answer 216

changing the password for an existing database user.

Answer 217

connecting to a database instance

Answer 218

querying data from tables across databases

Answer 219

starting up a database instance

Answer 220

executing operating system (OS) commands in a session

Answer 221

GRANT CREATE VIEW ON table1 TO user1;

Answer 222

GRANT ALTER TABLE TO PUBLIC;

Answer 223

GRANT CREATE TABLE TO user1, user2;

Answer 224

GRANT CREATE SESSION TO ALL;

Answer 225

INSERT INTO (SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_total FROM orders o, customers c WHERE o.customer_id = c.customer_id AND c.cust_last_name='Roberts' AND c.credit_limit=600) VALUES (1,'10-mar-2007', 'direct', (SELECT customer_id FROM customers WHERE cust_last_name='Roberts' AND credit_limit=600), 1000);

Answer 226

INSERT INTO orders (order_id, order_date, order_mode, (SELECT customer id FROM customers WHERE cust_last_name='Roberts' AND credit_limit=600), order_total); VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);

Answer 227

INSERT INTO orders VALUES (1,'10-mar-2007', 'direct', (SELECT customer_id FROM customers WHERE cust_last_name='Roberts' AND credit_limit=600), 1000);

Answer 228

INSERT INTO orders (order_id, order_date, order_mode, (SELECT customer_id FROM customers WHERE cust_last_name='Roberts' AND credit_limit=600), order_total); VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);

Answer 229

The table prefix is missing for the column names in the SELECT clause.

Answer 230

The NATURAL JOIN clause is missing the USING clause.

Answer 231

The DEPARTMENTS table is not used before the EMPLOYEES table in the FROM clause.

Answer 232

The EMPLOYEES and DEPARTMENTS tables have more than one column with the same column name and data type.

Answer 233

When the MAXVALUE limit for the sequence is reached, you can increase the MAXVALUE limit by using the ALTER SEQUENCE statement.

Answer 234

DELETE <sequencename> would remove a sequence from the database.

Answer 235

The numbers generated by a sequence can be used only for one table.

Answer 236

CURRVAL is used to refer to the last sequence number that has been generated.

Answer 237

When a database instance shuts down abnormally, the sequence numbers that have been cached but not used would be available once again when the database instance is restarted.

Answer 238

finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers

Answer 239

finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'

Answer 240

listing of customers who do not have a credit limit and were born before 1980

Answer 241

finding the number of customers, in each city, who’s marital status is 'married'.

Answer 242

listing of those customers, whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'.

Answer 243

A set of Data Manipulation Language (DML) statements executed in a sequence ending with a SAVEPOINT forms a single transaction.

Answer 244

Each Data Definition Language (DDL) statement executed forms a single transaction.

Answer 245

A set of DDL statements executed in a sequence ending with a COMMIT forms a single transaction.

Answer 246

A combination of DDL and DML statements executed in a sequence ending with a COMMIT forms a single transaction.

Answer 247

CREATE OR REPLACE VIEW ord_vu AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id) FROM orders o JOIN order_items i ON (o.order_id = i.order_id) GROUP BY o.order_id, o.order_date;

Answer 248

CREATE OR REPLACE VIEW ord_vu (order_id, order_date) AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id) "NO OF ITEMS" FROM orders o JOIN order_items i ON (o.order_id = i.order_id) GROUP BY o.order_id, o.order_date;

Answer 249

. CREATE OR REPLACE VIEW ord_vu AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id) "NO OF ITEMS" FROM orders o JOIN order_items i ON (o.order_id = i.order_id) GROUP BY o.order_id, o.order_date;

Answer 250

CREATE OR REPLACE VIEW ord_vu AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id) || "NO OF ITEMS" FROM orders o JOIN order_items i ON (o.order_id = i.order_id) WHITH CHECK OPTION;

Answer 251

It must have primary-key and foreign-key constraints defined on the columns used in the join condition.

Answer 252

It requires the column names to be the same in all tables used for the join conditions.

Answer 253

It is applicable for equijoin and nonequijoin conditions.

Answer 254

It is applicable for only equijoin conditions.

Answer 255

The names of columns in all SELECT statements must be identical.

Answer 256

It ignores NULL values.

Answer 257

Reversing the order of the intersected tables alters the result.

Answer 258

The number of columns and data types must be identical for all SELECT statements in the query.

Answer 259

It displays 5 percent of the products with the highest amount sold.

Answer 260

It displays the first 5 percent of the rows from the SALES table.

Answer 261

It displays 5 percent of the products with the lowest amount sold.

Answer 262

It results in an error because the ORDER BY clause should be the last clause.

	Creado por Jean Krüger hace alrededor de 7 años

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Simulado Prova 1Z0-071: Oracle Database OCA 12c

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