Evaluate the following code： SQLVARIABLE task_name VARCHAR2（255）； SQLVARIABLE sql_stmt VARCHAR2（4000）； SQLBEGIN ：sql_stmt ：= ’SELECT COUNT（*） FROM customers WHERE cust_state_province =’’CA’’’；：task_name ：= ’MY_QUICKTUNE_TASK’； DBMS_ADVISOR.QUICK_TUNE（DBMS_ADVISOR.SQLACCESS_ADVISOR，：task_name，：sql_stmt）； END； What is the outcome of this block of code？（） A、 It creates a task and workload， and executes the task.B、 It creates a task and workload but does not execute the task.C、 It produces an error because a template has not been created.D、 It produces an error because the SQL Tuning Set has not been created.

考题 The CUSTOMERS table has these columns:CUSTOMER_ID NUMBER (4) NOT NULLCUSTOMER_NAME VARCHAR2 (100) NOT NULLSTREET_ADDRESS VARCHAR2 (150)CITY_ADDRESS VARHCAR2 (50)STATE_ADDRESS VARCHAR2 (50)PROVINCE_ADDRESS VARCHAR2 (50)COUNTRY_ADDRESS VARCHAR2 (50)POSTAL_CODE VARCHAR2 (12)CUSTOMER_PHONE VARCHAR2 (20)The CUSTOMER_ID column is the primary key for the table.You need to determine how dispersed your customer base is.Which expression finds the number of different countries represented in the CUSTOMERS table?（）A. COUNT(UPPER(country_address))B. COUNT(DIFF(UPPER(country_address)))C. COUNT(UNIQUE(UPPER(country_address)))D. COUNT DISTINTC UPPER(country_address)E. COUNT(DISTINTC (UPPER(country_address)))

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考题 You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns:CUST_ID NUMBER(4) NOT NULLCUST_NAME VARCHAR2(100) NOT NULLCUST_ADDRESS VARCHAR2(150)CUST_PHONE VARCHAR2(20)Which SELECT statement accomplishes this task?（）A. SELECT* FROM customers;B. SELECT name, address FROM customers;C. SELECT id, name, address, phone FROM customers;D. SELECT cust_name, cust_address FROM customers;E. SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

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考题 Evaluatethefollowingcode：SQLVARIABLEtask_nameVARCHAR2（255）；SQLVARIABLEsql_stmtVARCHAR2（4000）；SQLBEGIN：sql_stmt：=’SELECTCOUNT（*）FROMcustomersWHEREcust_state_province=’’CA’’’；：task_name：=’MY_QUICKTUNE_TASK’；DBMS_ADVISOR.QUICK_TUNE（DBMS_ADVISOR.SQLACCESS_ADVISOR，：task_name，：sql_stmt）；END；Whatistheoutcomeofthisblockofcode？（）A.Itcreatesataskandworkload，andexecutesthetask.B.Itcreatesataskandworkloadbutdoesnotexecutethetask.C.Itproducesanerrorbecauseatemplatehasnotbeencreated.D.ItproducesanerrorbecausetheSQLTuningSethasnotbeencreated.

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考题 The CUSTOMERS table has these columns:CUSTOMER_ID NUMBER(4) NOT NULLCUSTOMER_NAME VARCHAR2(100) NOT NULLSTREET_ADDRESS VARCHAR2(150)CITY_ADDRESS VARCHAR2(50)STATE_ADDRESS VARCHAR2(50)PROVINCE_ADDRESS VARCHAR2(50)COUNTRY_ADDRESS VARCHAR2(50)POSTAL_CODE VARCHAR2(12)CUSTOMER_PHONE VARCHAR2(20)The CUSTOMER_ID column is the primary key for the table.You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?（）A.COUNT(UPPER(country_address))B.COUNT(DIFF(UPPER(country_address)))C.COUNT(UNIQUE(UPPER(country_address)))D.COUNT DISTINCT UPPER(country_address)E.COUNT(DISTINCT (UPPER(country_address)))

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考题 “雇员”表包含以下列：姓氏VARCHAR2（25）名字VARCHAR2（25）电子邮件VARCHAR2（50）如果要编写以下SELECT语句来检索具有电子邮件地址的雇员的姓名。SELECT姓氏||’，’||名字"雇员姓名"FROM雇员；则应使用哪条WHERE子句来完成此条语句？（）A、WHERE电子邮件=NULL；B、WHERE电子邮件!=NULL；C、WHERE电子邮件IS NULL；D、WHERE电子邮件IS NOT NULL；

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考题 The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which two statements find the number of customers? （） A、SELECT TOTAL(*) FROM customer;B、SELECT COUNT(*) FROM customer;C、SELECT TOTAL(customer_id) FROM customer;D、SELECT COUNT(customer_id) FROM customer;E、SELECT COUNT(customers) FROM customer;F、SELECT TOTAL(customer_name) FROM customer;

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考题 You accepted the recommended SQL Profile by executing the following code： DECLARE sqlprofile_name varchar2（30）； BEGIN sqlprofile_name ：= DBMS_SQLTUNE.ACCEPT_SQL_PROFILE（ task_name = ’my_task’， profile_name = ’my_profile’）； END； Which advisor will analyze this profile？（） A、 SQL Access AdvisorB、 Undo AdvisorC、 Segment AdvisorD、 SQL Tuning Advisor

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考题 Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith?（）A、SELECT COUNT(*) FROM employees WHERE last_name='Smith';B、SELECT COUNT(dept_id) FROM employees WHERE last_name='Smith';C、SELECT DISTINCT(COUNT(dept_id)) FROM employees WHERE last_name='Smith';D、SELECT COUNT(DISTINCT dept_id) FROM employees WHERE last_name='Smith';E、SELECT UNIQUE(dept_id) FROM employees WHERE last_name='Smith';

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考题 Examine the structure of the EMPLOYEES table: EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) DEPARTMENT_ID NUMBER SALARY NUMBER What is the correct syntax for an inline view? （）A、SELECT a.last_name, a.salary, a.department_id, b.maxsal FROM employees a, (SELECT department_id, max(salary)maxsal FROM employees GROUP BY department_id) b WHERE a.department_id = b.department_id AND a.salary b.maxsal;B、SELECT a.last name, a.salary, a.department_id FROM employees a WHERE a.department_id IN (SELECT department_id FROM employees b GROUP BY department_id having salary = (SELECT max(salary) from employees))C、SELECT a.last_name, a.salary, a.department_id FROM employees a WHERE a.salary = (SELECT max(salary) FROM employees b WHERE a.department _ id = b.department _ id);D、SELECT a.last_name, a.salary, a.department_id FROM employees a WHERE (a.department_id, a.salary) IN (SELECT department_id, a.salary) IN (SELECT department_id max(salary) FROM employees b GROUP BY department_id ORDER BY department _ id);

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考题 Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables: EMPLOYEES EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE NEW_EMPLOYEES EMPLOYEE_ID NUMBER Primary Key NAME VARCHAR2 (60) Which DELETE statement is valid?（）A、DELETE FROM employees WHERE employee_id = (SELECT employee_id FROM employees);B、DELETE * FROM employees WHERE employee_id = (SELECT employee_id FROM new_ employees);C、DELETE FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name = 'carrey');D、DELETE * FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name = 'carrey');

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考题 The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) A promotional sale is being advertised to the customers in France. Which WHERE clause identifies customers that are located in France?（）A、WHERE lower(country_address) = "france"B、WHERE lower(country_address) = 'france'C、WHERE lower(country_address) IS 'france'D、WHERE lower(country_address) = '%france%'E、WHERE lower(country_address) LIKE %france%

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考题 The EMPLOYEES table contains these columns: EMPLOYEE_ID NUMBER(4) ENAME VARCHAR2 (25) JOB_ID VARCHAR2(10) Which SQL statement will return the ENAME, length of the ENAME, and the numeric position of the letter "a" in the ENAME column, for those employees whose ENAME ends with a the letter "n"?（）A、SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, 'a') FROM EMPLOYEES WHERE SUBSTR (ENAME, -1,1) = 'n';B、SELECT ENAME, LENGTH(ENAME), INSTR(ENAME, ,-1,1) FROM EMPLOYEES WHERE SUBSTR (ENAME, -1,1) = 'n';C、SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR (ENAME, 1,1) = 'n';D、SELECT ENAME, LENGTH(ENAME), SUBSTR(ENAME, -1,1) FROM EMPLOYEES WHERE INSTR (ENAME, -1,1) = 'n';

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考题 The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) Which statement finds the rows in the CUSTOMERS table that do not have a postal code? （）A、SELECT customer_id, customer_name FROM customers WHERE postal_code CONTAINS NULL;B、SELECT customer_id, customer_name FROM customers WHER postal_code = ' ___________';C、SELECT customer_id, customer_name FROM customers WHERE postal _ code IS NULL;D、SELECT customer_id, customer_name FROM customers WHERE postal code IS NVL;E、SELECT customer_id, customer_name FROM customers WHERE postal_code = NULL;

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考题 Examine the description of the CUSTOMERS table: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which statement returns the city address and the number of customers in the cities Los Angeles or San Francisco?（）A、SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco');B、SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address;C、SELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address, customer_id;D、SELECT city_address, COUNT(customer_id) FROM customers GROUP BY city_address IN ('Los Angeles', 'San Francisco');

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考题 The EMP table has these columns: ENAME VARCHAR2(35) SALARY NUMBER(8,2) HIRE_DATE DATE Management wants a list of names of employees who have been with the company for more than five years. Which SQL statement displays the required results? （）A、SELECT ENAME FROM EMP WHERE SYSDATE-HIRE_DATE 5;B、SELECT ENAME FROM EMP WHERE HIRE_DATE-SYSDATE 5;C、SELECT ENAME FROM EMP WHERE (SYSDATE_HIRE_DATE)/365 5;D、SELECT ENAME FROM EMP WHERE (SYSDATE_HIRE_DATE)*/365 5;

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考题 The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER (4) NOT NULL CUSTOMER_NAME VARCHAR2 (100) NOT NULL STREET_ADDRESS VARCHAR2 (150) CITY_ADDRESS VARHCAR2 (50) STATE_ADDRESS VARCHAR2 (50) PROVINCE_ADDRESS VARCHAR2 (50) COUNTRY_ADDRESS VARCHAR2 (50) POSTAL_CODE VARCHAR2 (12) CUSTOMER_PHONE VARCHAR2 (20) The CUSTOMER_ID column is the primary key for the table. You need to determine how dispersed your customer base is. Which expression finds the number of different countries represented in the CUSTOMERS table?（）A、COUNT(UPPER(country_address))B、COUNT(DIFF(UPPER(country_address)))C、COUNT(UNIQUE(UPPER(country_address)))D、COUNT DISTINTC UPPER(country_address)E、COUNT(DISTINTC (UPPER(country_address)))

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考题 Examine the structure of the EMPLOYEES and NEW_EMPLOYEES tables: EMPLOYEES EMPLOYEE_ID NUMBER Primary Key FIRST_NAME VARCHAR2(25) LAST_NAME VARCHAR2(25) HIRE_DATE DATE NEW_EMPLOYEES EMPLOYEE_ID NUMBER Primary Key NAME VARCHAR2(60) Which DELETE statement is valid?（）A、DELETE FROM employees WHERE employee_id = (SELECT employee_id FROM employees);B、DELETE * FROM employees WHERE employee_id = (SELECT employee_id FROM new_employees);C、DELETE FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE name ='Carrey');D、DELETE * FROM employees WHERE employee_id IN (SELECT employee_id FROM new_employees WHERE last_name ='Carrey');

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考题单选题FACULTY表包含以下各列： FACULTYID VARCHAR2(5) NOT NULL PRIMARY KEY FIRST_NAME VARCHAR2(20) LAST_NAME VARCHAR2(20) ADDRESS VARCHAR2(35) CITY VARCHAR2(15) STATE VARCHAR2(2) ZIP NUMBER(9) TELEPHONE NUMBER(10) STATUS VARCHAR2(2) NOT NULL COURSE 表包含以下各列： COURSEID VARCHAR2(5) NOT NULL PRIMARY KEY SUBJECT VARCHAR2(5) TERM VARCHAR2(6 FACULTYID VARCHAR2(5) NOT NULL FOREIGN KEY 您需要制定一个报表，用于确定在下学期任教的所有副教授。您要创建一个视图来简化报表的创建过程。以下哪条CREATE VIEW语句将完成此任务（）A CREATE VIEW（SELECT first_name，last_name，status，courseid，subject，term FROM faculty，course WHERE facultyid=facultyid）B CREATE VIEW pt_view ON（SELEC Tfirst_name，last_name，status，courseid，subject，term FROM faculty f and coursec WHERE f.facultyid=c.facultyid）C CREATE VIEW pt_view IN（SELECT first_name，last_name，status，courseid，subject，term FROM faculty course）D CREATE VIEW pt_view AS（SELECT first_name，last_name，status，courseid，subject，term FROM facultyf，coursec WHERE f.facultyid=c.facultyid）

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考题单选题You accepted the recommended SQL Profile by executing the following code： DECLARE sqlprofile_name varchar2（30）； BEGIN sqlprofile_name ：= DBMS_SQLTUNE.ACCEPT_SQL_PROFILE（ task_name = ’my_task’， profile_name = ’my_profile’）； END； Which advisor will analyze this profile？（）A SQL Access AdvisorB Undo AdvisorC Segment AdvisorD SQL Tuning Advisor

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考题单选题Examine the description of the CUSTOMERS table: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which statement returns the city address and the number of customers in the cities Los Angeles or San Francisco?（）A SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco');B SELECT city_address, COUNT(*) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address;C SELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN ('Los Angeles', 'San Francisco') GROUP BY city_address, customer_id;D SELECT city_address, COUNT(customer_id) FROM customers GROUP BY city_address IN ('Los Angeles', 'San Francisco');

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考题多选题The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) The CUSTOMER_ID column is the primary key for the table. Which two statements find the number of customers?（）ASELECT TOTAL(*) FROM customer;BSELECT COUNT(*) FROM customer;CSELECT TOTAL(customer_id) FROM customer;DSELECT COUNT(customer_id) FROM customer;ESELECT COUNT(customers) FROM customer;FSELECT TOTAL(customer_name) FROM customer;

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考题单选题You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns: CUST_ID NUMBER(4) NOT NULL CUST_NAME VARCHAR2(100) NOT NULL CUST_ADDRESS VARCHAR2(150) CUST_PHONE VARCHAR2(20) Which SELECT statement accomplishes this task?（）A SELECT* FROM customers;B SELECT name, address FROM customers;C SELECT id, name, address, phone FROM customers;D SELECT cust_name, cust_address FROM customers;E SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

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考题单选题The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) A promotional sale is being advertised to the customers in France. Which WHERE clause identifies customers that are located in France?（）A WHERE lower(country_address) = franceB WHERE lower(country_address) = 'france'C WHERE lower(country_address) IS 'france'D WHERE lower(country_address) = '%france%'E WHERE lower(country_address) LIKE %france%

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考题单选题You need to produce a report for mailing labels for all customers. The mailing label must have only the customer name and address. The CUSTOMERS table has these columns: CUST_ID NUMBER(4) NOT NULL CUST_NAME VARCHAR2(100) NOT NULL CUST_ADDRESS VARCHAR2(150) CUST_PHONE VARCHAR2(20) Which SELECT statement accomplishes this task?（）A SELECT* FROM customers;B SELECT name, address FROM customers;C SELECT id, name, address, phone FROM customers;D SELECT cust_name, cust_address FROM customers;E SELECT cust_id, cust_name, cust_address, cust_phone FROM customers;

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考题单选题The CUSTOMERS table has these columns: CUSTOMER_ID NUMBER(4) NOT NULL CUSTOMER_NAME VARCHAR2(100) NOT NULL STREET_ADDRESS VARCHAR2(150) CITY_ADDRESS VARCHAR2(50) STATE_ADDRESS VARCHAR2(50) PROVINCE_ADDRESS VARCHAR2(50) COUNTRY_ADDRESS VARCHAR2(50) POSTAL_CODE VARCHAR2(12) CUSTOMER_PHONE VARCHAR2(20) Which statement finds the rows in the CUSTOMERS table that do not have a postal code? （）A SELECT customer_id, customer_name FROM customers WHERE postal_code CONTAINS NULL;B SELECT customer_id, customer_name FROM customers WHER postal_code = ' ___________';C SELECT customer_id, customer_name FROM customers WHERE postal _ code IS NULL;D SELECT customer_id, customer_name FROM customers WHERE postal code IS NVL;E SELECT customer_id, customer_name FROM customers WHERE postal_code = NULL;

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考题单选题Evaluate the following code： SQLVARIABLE task_name VARCHAR2（255）； SQLVARIABLE sql_stmt VARCHAR2（4000）； SQLBEGIN ：sql_stmt ：= ’SELECT COUNT（*） FROM customers WHERE cust_state_province =’’CA’’’；：task_name ：= ’MY_QUICKTUNE_TASK’； DBMS_ADVISOR.QUICK_TUNE（DBMS_ADVISOR.SQLACCESS_ADVISOR，：task_name，：sql_stmt）； END； What is the outcome of this block of code？（）A It creates a task and workload， and executes the task.B It creates a task and workload but does not execute the task.C It produces an error because a template has not been created.D It produces an error because the SQL Tuning Set has not been created.

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考题单选题Examine the description of the EMPLOYEES table: EMP_ID NUMBER(4) NOT NULL LAST_NAME VARCHAR2(30) NOT NULL FIRST_NAME VARCHAR2(30) DEPT_ID NUMBER(2) Which statement produces the number of different departments that have employees with last name Smith?（）A SELECT COUNT (*) FROM employees WHERE last _name='smith';B SELECT COUNT (dept_id) FROM employees WHERE last _name='smith';C SELECT DISTINCT (COUNT (dept_id) FROM employees WHERE last _name='smith';D SELECT COUNT (DISTINCT dept_id) FROM employees WHERE last _name='smith';E SELECT UNIQE (dept_id) FROM employees WHERE last _name='smith';

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