1z1-071 PDF Guide, Real 1z1-071 Exam

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Oracle 1z1-071 Exam Syllabus Topics:

Section	Objectives
Relational Database Concepts	- Tables, rows, columns, and relational model basics - Data types and database objects overview
Subqueries and Advanced SQL	- Single-row and multi-row subqueries - Correlated subqueries and nested queries
Data Manipulation and Schema Objects	- DML operations 1. INSERT, UPDATE, DELETE, MERGE - DDL operations 1. CREATE, ALTER, DROP tables and constraints
SQL Functions	- Single-row functions 1. Character, numeric, date functions - Conversion and conditional functions 1. TO_CHAR, TO_NUMBER, CASE expressions
SQL Retrieval	- Joins and set operations 1. INNER, OUTER, CROSS joins 2. UNION, INTERSECT, MINUS - SELECT statements and filtering data 1. WHERE clause and conditions 2. ORDER BY and row limiting

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Oracle Database SQL Sample Questions (Q158-Q163):

NEW QUESTION # 158
The customers table has the following structure:
You need to write a query that does the following tasks:
1. Display the first name and tax amount of the customers. Tax is 5% of their credit limit.
2. Only those customers whose income level has a value should be considered.
3. Customers whose tax amount is null should not be considered.
Which statement accomplishes all the required tasks?

A. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE cust_income_level <> NULL ANDtax_amount <> NULL;
B. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE (cust_income_level, tax_amount) IS NOT NULL;
C. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE cust_income_level IS NOT NULL ANDtax_amount IS NOT NULL;
D. SELECT cust_first_name, cust_credit_limit * .05 AS TAX_AMOUNTFROM customersWHERE cust_income_level IS NOT NULL ANDcust_credit_limit IS NOT NULL;

Answer: D

NEW QUESTION # 159

Which two queries only return CUBE?

A. SELECT shape FROM bricks JOIN boxes ON weight > min_weight;
B. SELECT shape FROM bricks JOIN boxes ON weight >= min_weight AND weight < max_weight;
C. SELECT shape FROM bricks JOIN boxes ON NOT (weight > max_weight);
D. SELECT shape FROM bricks JOIN boxes ON weight BETWEEN min_weight AND max_weight;
E. SELECT shape FROM bricks JOIN boxes ON weight < max_weight;

Answer: B,C

Explanation:
Based on the table structure given in the image, to return the value 'CUBE' from the 'bricks' table when joined with 'boxes', the condition must ensure that the weight of the bricks is within the allowed weight range specified in the 'boxes' table for a 'SMALL' box size.
A). True. Since MAX_WEIGHT is 0, a comparison using >= min_weight AND weight < max_weight will only return rows where the weight is less than 0, which is impossible for actual weight values, suggesting there might be a mistake in the data provided or the comparison logic.
E). True. NOT (weight > max_weight) effectively translates to 'where weight is less than or equal to max_weight'. However, since MAX_WEIGHT is 0, this condition would only be true if the weight is not greater than 0, which can only happen if the weight is 0 or less. This seems to indicate an anomaly where either the data is incorrect, or the condition is meant to handle a case where the weight is zero or possibly a negative placeholder value.
Both B and D will potentially return more than just 'CUBE' if there are bricks with weights greater than MIN_WEIGHT. C is incorrect because BETWEEN is inclusive, and there are no weights that are both greater than or equal to MIN_WEIGHT and less than or equal to MAX_WEIGHT when MAX_WEIGHT is
0.

NEW QUESTION # 160
You want to write a query that prompts for two column names and the WHEREcondition each time it is executed in a session but only prompts for the table name the first time it is executed.
The variables used in your query are never undefined in your session.
Which query can be used?

Answer: E

NEW QUESTION # 161
View the exhibit and examine the structure of ORDERSand CUSTOMERStables.

Which INSERT statement should be used to add a row into the ORDERStable for the customer whose CUST_LAST_NAMEis Robertsand CREDIT_LIMITis 600? Assume there exists only one row with CUST_LAST_NAME as Roberts and CREDIT_LIMIT as 600.

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

NEW QUESTION # 162
View the Exhibit and examine the ORDERS table.
The ORDERS table contains data and all orders have been assigned a customer ID. Which statement would add a NOT NULL constraint to the CUSTOMER_ID column?

A. ALTER TABLE orders
MODIFY customer_id CONSTRAINT orders_cust_id_nn NOT NULL;
B. ALTER TABLE orders
MODIFY CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);
C. ALTER TABLE orders
ADD customer_id NUMBER(6)CONSTRAINT orders_cust_id_nn NOT NULL;
D. ALTER TABLE orders
ADD CONSTRAINT orders_cust_id_nn NOT NULL (customer_id);

Answer: A

NEW QUESTION # 163
......

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