varchar2最大多大（） A、1000B、2000C、3000D、4000
varchar2最大多大（）
- A、1000
- B、2000
- C、3000
- D、4000
相关考题：
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)))
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)))
（若oracle数据库不了解，可以用sql）test1ID Varchar2(6) 主键NAME Varchar2(15)Test2Use_ID Varchar2(6) 主键关联TEST1的主键的IDPASSword Varchar2(10)
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）
ORACLE中char类型与varchar2类型的区别，描述正确的是（）A、char为定长字符数据类型B、char为不定长字符数据类型C、varchar2为不定长数据类型D、char与varchar2没有区别E、varchar2为定长数据类型
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? （）ASELECT customer_id, customer_name FROM customers WHERE postal_code CONTAINS NULL;BSELECT customer_id, customer_name FROM customers WHER postal_code = ' ___________';CSELECT customer_id, customer_name FROM customers WHERE postal _ code IS NULL;DSELECT customer_id, customer_name FROM customers WHERE postal code IS NVL;ESELECT customer_id, customer_name FROM customers WHERE postal_code = NULL;
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 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?（）ACOUNT(UPPER(country_address))BCOUNT(DIFF(UPPER(country_address)))CCOUNT(UNIQUE(UPPER(country_address)))DCOUNT DISTINTC UPPER(country_address)ECOUNT(DISTINTC (UPPER(country_address)))
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;
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?（）ASELECT city_address, COUNT(*) FROM customers WHERE city _ address IN ('Los Angeles','San Fransisco');BSELECT city_address, COUNT (*) FROM customers WHERE city address IN ( 'Los Angeles', 'San Fransisco') GROUP BY city_address;CSELECT city_address, COUNT(customer_id) FROM customers WHERE city_address IN ( 'Los Angeles', 'San Fransisco') GROUP BY city_address, customer_ id;DSELECT city_address, COUNT (customer_id) FROM customers GROUP BY city_ address IN ('Los Angeles','San Fransisco');
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) 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?（）ACOUNT(UPPER(country_address))BCOUNT(DIFF(UPPER(country_address)))CCOUNT(UNIQUE(UPPER(country_address)))DCOUNT DISTINCT UPPER(country_address)ECOUNT(DISTINCT (UPPER(country_address)))