Oracle 1Z0-047 echte frage : Oracle Database SQL Expert

Überzeugen Sie sich selbst, bevor Sie kaufen: EchteFrage stellt zur 1Z0-047 Prüfung eine kostenlose PDF-Demo bereit. So prüfen Sie in Ruhe, ob die Übungsfragen zur Oracle Database SQL Expert Ihren Ansprüchen genügen.

Oracle 1Z0-047 Prüfungsübersicht:

Zertifizierungsanbieter:	Oracle
Prüfungsname:	Oracle Database SQL Expert
Prüfungsnummer:	1Z0-047
Prüfungsformat:	Einzelauswahl, Mehrfachauswahl
Verwandte Zertifizierungen:	Oracle Database Administration Certifications (OCA/OCP tracks) Oracle Database SQL Certified Associate
Verfügbare Sprachen:	Englisch
Prüfungsgebühr:	ca. 245 USD
Prüfungsdauer:	120 Minuten
Empfohlenes Training:	SQL-Schulung bei Oracle Learning
Prüfungsanmeldung:	Oracle-Prüfungen bei Pearson VUE Registrierung für Oracle-Zertifizierungen
Beispielfragen:
Prüfungsmethode:	Die Prüfung wird in Testzentren von Pearson VUE oder als online überwachte Prüfung abgelegt
Voraussetzungen:	Es gibt keine formellen Zulassungsvoraussetzungen; Grundkenntnisse im Bereich SQL und relationaler Datenbanken werden jedoch empfohlen.
Offizielle Syllabus-URL:	https://education.oracle.com/oracle-certification-exams-list

Oracle 1Z0-047 Prüfungsthemen:

1Z0-047 Prüfungs-FAQ: Ihre Fragen, unsere Antworten

Oracle Database SQL Expert 1Z0-047 Prüfungsfragen mit Lösungen

Frage #1

Which CREATE TABLE statement is valid?

A. CREATE TABLE ord_details
(ord_no NUMBER(2),
item_no NUMBER(3),
ord_date date DEFAULT SYSDATE NOT NULL,
CONSTRAINT ord_pk PRIMARY KEY (ord_no, item_no));
B. CREATE TABLE ord_details
(ord_no NUMBER(2) UNIQUE, NOT NULL,
item_no NUMBER(3),
ord_date date DEFAULT SYSDATE NOT NULL);
C. CREATE TABLE ord_details
(ord_no NUMBER(2) PRIMARY KEY,
item_no NUMBER(3) PRIMARY KEY,
ord_date date NOT NULL);
D. CREATE TABLE ord_details
(ord_no NUMBER(2) ,
item_no NUMBER(3),
ord_date date DEFAULT NOT NULL,
CONSTRAINT ord_uq UNIQUE (ord_no),
CONSTRAINT ord_pk PRIMARY KEY (ord_no));

Frage #2

View the Exhibit and examine the descriptions of ORDER_ITEMS and ORDERS tables.
You want to display the CUSTOMER_ID, PRODUCT_ID, and total (UNIT_PRICE multiplied by QUANTITY) for the order placed. You also want to display the subtotals for a CUSTOMER_ID as well as for a PRODUCT ID for the last six months.

A. SELECT o.customer_Id, oi.productj_id, SUM(oi.unit_price*oi. quantity) "Total"
FROM order_items oi JOIN orders o
ON oi.order_id=o.order_id
GROUP BY ROLLUP (o.customer_id.oi.product_id)
WHERE MONTHS_BETWEEN(order_date, SYSDATE) <= 6;
B. SELECT o.customer_id, oi.product_id, SUM(oi.unit_price*oi.quantity) "Total"
FROM order_items oi JOIN orders o
ON oi.order_id=o.order_id
WHERE MONTHS_BETWEEN(order_date, SYSDATE) <= 6
GROUP BY ROLLUP (o.customer_id, oi.product_id);
C. SELECT o.customer_id, oi.product_id, SUM(oi.unit_price*oi. quantity) "Total"
FROM orderjtems oi JOIN orders o
ON oi.order_id=o.order_id
GROUP BY ROLLUP (o.customer_id.oi.product_id)
HAVING MONTHS_BETWEEN(order_date, SYSDATE) <= 6;
D. SELECT o.customer_id, oi.product_id, SUM(oi.unit_price*oi.quantity) "Total"
FROM order_items oi JOIN orders o
ON oi.order_id=o.order_id
GROUP BY ROLLUP (o.customer_id, oi.product_id)
WHERE MONTHS_BETWEEN(order_date, SYSDATE) >= 6;

Frage #3

View the Exhibit and examine the description of the CUSTOMERS table. You want to add a constraint on the CUST_FIRST_NAME column of the CUSTOMERS table so
that the value inserted in the column does not have numbers. Which SOL statement would you use to accomplish the task?

A. ALTER TABLE CUSTOMERS ADD CONSTRAINT cust_f_name CHECK(REGEXP_LIKE(cust_first_name1'^A-Z '))NOVALIDATE;
B. ALTER TABLE CUSTOMERS ADD CONSTRAINT cust_f_name CHECK(REGEXP_LIKE(cust_first_name,'[[:digit: ]]'))NOVALIDATE ;
C. ALTER TABLE CUSTOMERS ADD CONSTRAINT cust_f_name CHECK(REGEXP_LIKE(cust_first_name,'^[0-9] '))NOVALIDATE;
D. ALTER TABLE CUSTOMERS ADD CONSTRAINT cust_f_name CHECK(REGEXP_LIKE(cust_first_name,'[[:alpha: ]]'))NOVALIDATE;

Frage #4

Which statement is true regarding the ROLLUP operator specified in the GROUP BY clause of a SQL statement?

A. It produces only the grand totals for the groups specified in the GROUP BY clause.
B. It produces higher-level subtotals, moving from right to left through the list of grouping columns specified in the GROUP BY clause.
C. It produces only the subtotals for the groups specified in the GROUP BY clause.
D. It produces higher-level subtotals, moving in all the directions through the list of grouping columns specified in the GROUP BY clause.

Frage #5

Evaluate the following SELECT statement and view the Exhibit to examine its output: SELECT constraint_name, constraint_type, search_condition, r_constraint_name, delete_rule, status FROM user_constraints WHERE table_name = ORDERS
Which two statements are true about the output? (Choose two.)

A. In the second column, indicates a check constraint.
B. The column DELETE_RULE decides the state of the related rows in the child table when the corresponding row is deleted from the parent table.
C. The STATUS column indicates whether the table is currently in use.
D. The R_CONSTRAINT_NAME column gives the alternative name for the constraint.

Fragen und Antworten: