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To prepare for this certification exam, candidates should have a good understanding of SQL programming concepts, such as data types, operators, functions, and control structures. They should also be familiar with basic database concepts, such as normalization, indexing, and transaction management. 1z1-071 exam consists of 73 multiple-choice questions, which need to be completed within 100 minutes. The passing score for 1z1-071 exam is 63%. Upon successful completion of the exam, candidates will be awarded the Oracle Database SQL certification, which is a valuable credential for anyone seeking to advance their career in database management.

The Oracle 1z1-071 Exam consists of 73 questions and has a time limit of 100 minutes. The questions are of multiple-choice format and cover a wide range of topics, including SQL concepts, data manipulation, data definition, and SQL queries. Candidates who pass the exam will receive a certification that demonstrates their proficiency in SQL and can help them advance their careers as database developers or administrators.

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Oracle Database SQL Sample Questions (Q155-Q160):

NEW QUESTION # 155
Examine the description of the PROMOTIONStable:

You want to display the unique promotion costs in each promotion category.
Which two queries can be used? (Choose two.)

A. ORDER BY 1;
SELECT promo_cost, promo_category FROM promotions ORDER BY 1;
B. SELECT promo_category, DISTINCT promo_cost FROM promotions ORDER BY 2;
C. promotions ORDER BY 1;
SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1;
D. SELECT DISTINCT promo_cost ||' in '|| DISTINCT promo_category FROM promotions
E. SELECT DISTINCT promo_category || ' has ' || promo_cost AS COSTS FROM

Answer: B,C

NEW QUESTION # 156
Which three statements are true regarding indexes?

A. An update to a table can result in updates to any or all of the table's indexes.
B. An update to a table can result in no updates to any of the table's indexes.
C. A SELECT statement can access one or more indices without accessing any tables.
D. A UNIQUE index can be altered to be non-unique.
E. When a table is dropped and is moved to the RECYCLE BIN, all Indexes built on that table are permanently dropped.
F. A table belonging to one user can have an index that belongs to a different user,

Answer: B,E,F

NEW QUESTION # 157
Examine the data in the CUST NAME column of the CUSTOMERS table:
CUST_NAME
------------------------------
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You want to display the CUST_NAME values where the last name starts with Mc or MC. Which two WHERE clauses give the required result?

A. WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,'') +1)) LIKE'Mc%'
B. WHERE UPPER (SUBSTR(cust_name, INSTR(cust_name, '') +1)) LIKE UPPER('MC%')
C. WHERE SUBSTR(cust_name, INSTR(cust_name,'') +1) LIKE'Mc%'
D. WHERE INITCAP (SUBSTR(cust_name, INSTR(cust_name,'') +1)) IN ('MC%','Mc%)
E. WHERE SUBSTR(cust_name,INSTR(cust_name,'') +1) LIKE'Mc%' OR'MC%'

Answer: A,B

Explanation:
To find customers whose last names start with "Mc" or "MC", we need to ensure our SQL query correctly identifies and compares these prefixes regardless of case variations. Let's analyze the given options:
Option B: WHERE UPPER(SUBSTR(cust_name, INSTR(cust_name, ' ') + 1)) LIKE UPPER('MC%') This clause uses UPPER to convert both the extracted substring (starting just after the first space, assuming it indicates the start of the last name) and the comparison string 'MC%' to uppercase. This ensures case-insensitive comparison. The LIKE operator is used to match any last names starting with "MC", which will correctly capture both "Mc" and "MC". This option is correct.
Option C: WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name, ' ') + 1)) LIKE 'Mc%' This clause applies INITCAP to the substring, which capitalizes the first letter of each word and makes other letters lowercase. The result is compared to 'Mc%', assuming only the last name follows the space. This approach will match last names starting with "Mc" (like "McEwen"), but not "MC". However, considering we're looking for "Mc" specifically, this clause works under the assumption that "Mc" is treated as proper capitalization for these last names. Thus, it can also be considered correct, albeit less inclusive than option B.
The other options either use incorrect syntax or apply case-sensitive matches without ensuring that both "Mc" and "MC" are captured:
Option A: Contains syntax errors (unmatched quotes and wrong use of IN).
Option D: Uses case-sensitive match without combining both "Mc" and "MC".
Option E: Only matches "Mc", which is too specific.

NEW QUESTION # 158
Examine the structure of the INVOICEtable.
Name Null? Type
- ----------------- ------------------- -------------
INV_NO NOT NULL NUMBER(3)
INV_DATE DATE
INV_AMT NUMBER(10,2)
Which two SQL statements would execute successfully?

A. SELECT inv_no, NVL2(inv_amt, inv_date, 'Not Available')
FROM invoice;
B. SELECT inv_no, NVL2(inv_date, sysdate-inv_date, sysdate)
FROM invoice;
C. SELECT inv_no, NVL2(inv_amt, inv_amt*.25, 'Not Available')
FROM invoice;
D. SELECT inv_no, NVL2(inv_date, 'Pending', 'Incomplete')
FROM invoice;

Answer: B,D

NEW QUESTION # 159
View the exhibit and examine the data in the PROJ_TASK_DETAILS table. (Choose the best answer.) The PROJ_TASK_DETAILS table stores information about project tasks and the relation between them.
The BASED_ON column indicates dependencies between tasks.
Some tasks do not depend on the completion of other tasks.
You must generate a report listing all task IDs, the task ID of any task upon which it depends and the name of the employee in charge of the task upon which it depends.
Which query would give the required result?

A. SELECT p.task_id, p.based_on, d.task_in_chargeFROM proj_task_details p JOIN proj_task_details dON (p.based_on = d.task_id);
B. SELECT p.task_id, p.based_on, d.task_in_chargeFROM proj_task_details p JOIN proj_task_details dON (p.task_id = d.task_id);
C. SELECT p.task_id, p.based_on, d.task_in_chargeFROM proj_task_details p FULL OUTER JOIN proj_task_details dON (p.based_on = d.task_id);
D. SELECT p.task_id, p.based_on, d.task_in_chargeFROM proj_task_details p LEFT OUTER JOIN proj_task_details dON (p.based_on = d.task_id);

Answer: D

NEW QUESTION # 160
......

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1z1-071 Customized Lab Simulation Free PDF | Professional 1z1-071 Valid Braindumps Pdf: Oracle Database SQL

1z1-071 Valid Braindumps Pdf & 1z1-071 Certification Practice

Oracle Database SQL Sample Questions (Q155-Q160):

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