Display Customer name (concatenate First Name and Last Name with a space between) and total number of orders for each customer. Name the total number of orders column as 'NumberOfOrders'. Sort based on Number of orders in descending order
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Display Customer name (concatenate First Name and Last Name with a space between) and total number of orders for each customer. Name the total number of orders column as 'NumberOfOrders'. Sort based on Number of orders in descending order.
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- My resualt has error? SELECT last_name,customer_number2 from CUSTOMERS WHERE REFERRED is not null;SELECT last_name,customer_number*ERROR at line 1:ORA-00904: "CUSTOMER_NUMBER": invalid identifier List all customers who were referred to the bookstore by another customer. List eachcustomer’s last name and the number of the customer who made the referral.The InstantRide User Satisfaction team received a phone call from a user who might have left her wallet in the car. She indicated that the license plate of the car was starting with BB-883 but unfortunately, she could not remember the full license plate number. The team wants to get all travel information for the cars with the license plate starting with BB-883. You need to return all travel data from the TRAVELS table for the CAR_ID which has a plate number compared with SUBSTR and UPPER functions.Assuming the HOMEWORK10 table has three columns (Col1, Col2, and Col3, in this order), which of the following commands stores a NULL value in Col3 of the HOMEWORK10 table?a. INSERT INTO homework10 VALUES (‘A’, ‘B’, ‘C’);b. INSERT INTO homework10 (col3, col1, col2) VALUES (NULL, ‘A’, ‘B’);c. INSERT INTO homework10 VALUES (NULL, ‘A’, ‘B’);d. UPDATE homework10 SET col1 = col3;
- The primary keys are identified below. The following data types are defined in the SQL Server. tblLevels Level – Identity PKClassName – text 20 – nulls are not allowed tblPoolPool – Identity PKPoolName – text 20 – nulls are not allowedLocation – text 30 tblStaffStaffID – Identity PKFirstName – text 20MiddleInitial – text 3LastName – text 30Suffix – text 3Salaried – BitPayAmount – money tblClassesLessonIndex – Identity PKLevel – Integer FKSectionID – IntegerSemester – TinyIntDays – text 20Time – datetime (formatted for time)Pool – Integer FKInstructor – Integer FKLimit – TinyIntEnrolled – TinyIntPrice – money tblEnrollmentLessonIndex – Integer FKSID – Integer FK (LessonIndex and SID) Primary KeyStatus – text 30Charged – bitAmountPaid – moneyDateEnrolled – datetime tblStudentsSID – Identity PKFirstName – text 20MiddleInitial – text 3LastName – text 30Suffix – text 3Birthday – datetimeLocalStreet – text 30LocalCity – text 20LocalPostalCode – text 6LocalPhone – text 10 Implement this…I need the output of the queries create table Accounts (account_no number primary key, client_name varchar2(200),Gender varchar2(20),address Varchar2(200)); create table loans (branch_id number, bank_name varchar2(200),Loan number, loan_Status varchar2(200),account_no number references Accounts(account_no) ON DELETE CASCADE); 1. delete from accounts where account_no= 101; 2. select * from accounts a where exists (select 1 from loans l where l.account_no= a.account_no) 3. select * from accounts a where exists (select 1 from loans l where l.account_no= a.account_no and status = 'accepted') 4. create or replace view view1 as select client_name , address,l.bank_name,l.loan_Status from accounts a, loans l where a.account_no = l.account_no Step 2 5.create or replace view view2 as select client_name , address,l.bank_name,l.loan_Status from accounts a, loans l where a.account_no = l.account_no and address = 'Karachi' 6. create or replace view view3 as select client_name ,…The view V_PAT_ADT_LOCATION_HX returns one row of information per bed stay. The column ADT_DEPARTMENT_ID is the ID of the bed's department. The column ADT_DEPARTMENT_NAME is that department's name. Two distinct departments can potentially share the same name. Which of the following queries would return exactly one row per department and display the name of the department? A. SELECT ADT_DEPARTMENT_ID, MIN(ADT_DEPARTMENT_NAME) FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID B. SELECT MIN(ADT_DEPARTMENT_ID), ADT_DEPARTMENT_NAME FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID C. SELECT 'ADT_DEPARTMENT_ID', ADT_DEPARTMENT_NAME FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID D. SELECT ADT_DEPARTMENT_ID, 'ADT_DEPARTMENT_NAME' FROM V_PAT_ADT_LOCATION_HX GROUP BY ADT_DEPARTMENT_ID
- CREATE TABLE test( test_id INTEGER PRIMARY KEY, test_name varchar2 (255) ,t_date DATE, t_result varchar2(255));ALTER TABLE test ADD (t_time varchar2(1));INSERT INTO test values ('5454','corona','monday','11:15PM','positive'); (6567,'corona','sunday','10:50AM','negative');what is these codes insert into values()... CREATE TABLE [Car] ([CarID] varchar(30),[SerialNumber] varchar(50),[Model] varchar(30),[Colour] varchar(30),[Year] varchar(20),PRIMARY KEY ([CarID])); CREATE TABLE [ServiceTicket] ([ServiceticketID] varchar(40),[Serviceticketnumber] varchar(40),[CarID] varchar(30),[CustomerID] int,[Datarecieved] varchar(30),[Comments] varchar(100),[Datareturnedcustomer] varchar(100),PRIMARY KEY ([ServiceticketID]),CONSTRAINT [FK_ServiceTicket.CustomerID]FOREIGN KEY ([CustomerID])REFERENCES [Customer]([CustomerID]),CONSTRAINT [FK_ServiceTicket.CarID]FOREIGN KEY ([CarID])REFERENCES [Car]([CarID])); CREATE TABLE [ServiceMechanic] ([ServicemechanicID] varchar(40),[ServiceticketID] varchar(40),[ServiceID] int,[MechanicID] int,[Hours] varchar(10),[Comment] varchar(100),[Rate] varchar(40),PRIMARY KEY ([ServicemechanicID]),CONSTRAINT [FK_ServiceMechanic.MechanicID]FOREIGN KEY ([MechanicID])REFERENCES [Mechanic]([MechanicID]),CONSTRAINT…Assume you have these table CREATE TABLE Student (studid Char(9) Primary Key,FirstName Varchar2(50) NOT NULL,LastName Varchar2(50) NOT NULL,Email Varchar2(50) NOT NULL UNIQUE,PhoneNumber Number(8) NOT NULL UNIQUE,DateOfBirth Date NOT NULL,GPA Number(1,2) NOT NULL); CREATE TABLE Club (clubID Number(3) Primary Key,ClubName Varchar2(20) NOT NULL,studid Char(9) NOT NULL UNIQUE, Foreign key(studid) references Student(studid)); CREATE TABLE MemberOf (ClubID Number(3) Primary Key, Studid char(9), JoiningDate date, Foreign Key(ClubID) references club(clubID), Foreign Key(Studid) references Student(studid)); CREATE TABLE Activities (Actid Number(3) Primary Key,Acdt DATE NOT NULL,Place Varchar2(50) NOT NULL,durationNbHour Number NOT NULL); CREATE TABLE Organize (actid Number(3) Primary Key, clubID Number(3), Fee Number(4,2), Foreign Key(actid) references Activities(Actid),Foreign Key(clubID) references MemberOf(ClubID)); Write in SQL queries to populate the above tables by inserting minimum 4…
- Add a connection string in the appsettings.json file.Open the Startup.cs file to add the SQL server db provider for EF Core. Add the code services.AddDbContext<ProductContext>(o => o.UseSqlServer(Configuration.GetConnectionString("ProductDB"))); under ConfigureServices method. Note that in the GetConnectionString method the name of the key of the connection string is passed that was added in appsettings file.Which operator is used to process a correlated subquery?a. EXISTSb. INc. LINKd. MERGEThe PatientDim table contains columns. PatientEpicId and name. Use the following query to answer the questions below it: SELECT PatientEpicID Name FROM PatientDim What does this query return? select one of the following: A. Nothing because the query won't run.B. The column PatientEpicId and name from the table PatientDimC. The PatientEpicID column from the table PatientDim, with an alias of name.D. The name column from the table PatientDim, with an alias of PatientEpicId