Tuesday, 10 March 2026

sql 1

 SQL Exercise 1


1


create table EmployeeName(

EmployeeID int identity(1,1) Primary key not null,

FirstName varchar(100) not null,

MiddleName varchar(100),

LastName varchar(100) not null,

CallOutName varchar(100),

LastUpdatedDate datetime default GetDATE()

)


create table EmployeeOfficialDetails(

OfficialID int identity(1,1) Primary key not null,

EmployeeID int null,

Designation varchar(100) null,

JoinedYear int not null,

ResignedYear int null,

LastUpdatedDate datetime default GetDATE()

)


alter table EmployeeOfficialDetails 

add constraint Fk_EmployeeOfficialDetails_EmployeeName

foreign key(EmployeeID)

references EmployeeName(EmployeeID)



create table EmployeeExperienceDetails(

ExperienceID int identity(1,1) Primary key not null,

OfficialID int null,

ReportTo varchar(100) null,

Location varchar(100) null,

LastUpdatedDate datetime default GetDATE(),

Constraint fk_EmployeeExperienceDetails_EmployeeOfficialDetails 

foreign key(OfficialID) references EmployeeOfficialDetails(OfficialID)

)


2 select * from Nurse where Registered = 0


3

select [Name], position 

from Nurse 

where Position = 'Head Nurse'


select d.Name as Department, p.Name as Physician 

from Department d

join Physician p

on d.Head = p.EmployeeID







5:

select count(distinct(Patient)) as 'Number of Patients taken at least one appointment'

from Appointment


Another way for more accurate


select count(distinct(ap.Patient)) as 'Number of Patients taken at least one appointment'

from Appointment ap

join Physician p

on ap.Physician = p.EmployeeID


6

select BlockFloor as [Floor], BlockCode as Code from Room

where Number = 212


7

select count(Number) as 'Number of available rooms' from Room

where Unavailable = 0


8

select count(Number) as 'Number of unavailable rooms' from Room

where Unavailable = 1







9

select p.Name as Physician, d.Name as Department

from Physician p

join Affiliated_With a

on p.EmployeeID = a.Physician

join Department d

on d.DepartmentID = a.Department

where a.PrimaryAffiliation = 1


10

select p.Name as Physician, pr.Name as Treatment

from Physician p

join Trained_In t on p.EmployeeID = t.Physician

join ProcedureTbl pr on pr.Code = t.Treatment


11

select p.Name as Physician, p.Position, d.Name as Department

from Physician p

join Affiliated_With a

on p.EmployeeID = a.Physician

join Department d

on d.DepartmentID = a.Department

where a.PrimaryAffiliation = 0


12

select p.Name as Physician, p.Position

from Physician p

left join Trained_In t on p.EmployeeID = t.Physician where t.Treatment is null


13

select pt.Name as Patient, pt.Address, ph.Name as Physician

from Patient pt

join Physician ph on pt.PCP = ph.EmployeeID


14

select p.Name as PatientName, COUNT(a.Physician) as 'Appointment for No. of Physicians'

from Appointment a

left join Patient p

on a.Patient = p.SSN

group by p.Name


15

select count(distinct(Patient)) as 'No. of patient got appointment for room C'

from Appointment

where ExaminationRoom = 'C'


16


select p.Name as Patient, a.ExaminationRoom as 'Room No', 

a.Start as 'Date and Time of appointment'

from Patient p

join Appointment a

on p.SSN = a.Patient




17

select n.Name as Nurse, a.ExaminationRoom as 'Room No.' from Nurse n

join Appointment a

on n.EmployeeID = a.PrepNurse


18

select p.Name as Patient,n.Name as 'Name of Nurse assisting the Physician', ph.Name as Physician, a.ExaminationRoom as 'Room No.', a.Start as 'Start_dt_time'

from Appointment a

join Patient p on a.Patient = p.SSN

join Physician ph on a.Physician = ph.EmployeeID

join Nurse n on a.PrepNurse = n.EmployeeID

where a.Start = '2008-04-25 10:00:00'

19

select p.Name as 'Name of Patient', ph.Name as 'Name of Physician', a.ExaminationRoom as 'Room No.'

from Appointment a

join Patient p on a.Patient = p.SSN

join Physician ph on a.Physician = ph.EmployeeID

where a.PrepNurse is null


20

select  p.Name as 'Patient', ph.Name as 'Physician', m.Name as 'Medication' 

from Patient p

join Physician ph on p.PCP = ph.EmployeeID

join Prescribes prb on prb.Patient = p.SSN

join Medication m on m.Code = prb.Medication



21


select  p.Name as 'Patient', ph.Name as 'Physician', m.Name as 'Medication'  

from Patient p

join Physician ph on p.PCP = ph.EmployeeID

join Prescribes prb on prb.Patient = p.SSN

join Medication m on m.Code = prb.Medication

where prb.Appointment is not null


22

select  p.Name as 'Patient', ph.Name as 'Physician', m.Name as 'Medication'  

from Patient p

join Physician ph on p.PCP = ph.EmployeeID

join Prescribes prb on prb.Patient = p.SSN

join Medication m on m.Code = prb.Medication

where prb.Appointment is null


23


select BlockCode as Block, count(*) as 'Number of available rooms' 

from Room

where Unavailable = 0

group by BlockCode

order by BlockCode





24


select BlockFloor as Block, count(*) as 'Number of available rooms' 

from Room

where Unavailable = 0

group by BlockFloor

order by BlockFloor


25


select BlockFloor as Floor, BlockCode as Block, count(*) as 'Number of available rooms' 

from Room

where Unavailable = 0

group by BlockFloor, BlockCode

order by BlockFloor


26


select BlockFloor as Floor, BlockCode as Block, count(*) as 'Number of unavailable rooms' 

from Room

where Unavailable = 1

group by BlockFloor, BlockCode

order by BlockFloor





27

select top 1 BlockFloor as Floor,  count(*) as 'Number of available rooms' 

from Room

where Unavailable = 0

group by BlockFloor

order by count(*) desc


28

select top 1 with ties BlockFloor as Floor,  count(*) as 'Number of available rooms' 

from Room

where Unavailable = 0

group by BlockFloor

order by count(*) asc


29

select p.Name as Patient, s.Room, r.BlockFloor as [Floor], r.BlockCode as Block 

from Patient p

join Stay s on s.Patient = p.SSN

join Room r on r.Number = s.Room


30

select n.Name as Nurse, o.BlockCode as [Block]

from Nurse n

join On_Call o on o.Nurse = n.EmployeeID


Monday, 23 June 2025

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SELECT * FROM Employee;








What will be output of below SQL Query  ?

SELECT (FirstName +' ' + LastName) as FullName
FROM Employee 
        WHERE EmployeeID = 1;


Answer : Error

Reason of error:
  • When using aliase, it must be a single world.
  • Alliase will not accept space

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Tuesday, 15 April 2025

LinkedIn How to delete past Resume applied for job ?

                  In LinkedIn How to manage Resume?
When you apply for Job, you upload a resume.
and each time if you have uploaded new resume then LinkedIn will save that resume.

Benefit:
Next time while applying for new job, you can reuse those save resume.

Issue:
While applying for new Job, if you wish to use those existing saved resumes then it is difficult to remember internal details of resume so which one as a candidate need to use.
It’s a better to remove those older resumes.

In LinkedIn, how to manage Resume?

1: How to delete saved resume (previous applied resume)

while applying for new Job?

 

2: How to add new resume on LinkedIn?

Solution:
Open LinkedIn

1.   Navigate to the Jobs tab.

2.   Click on Preferences.

3.   In the pop-up modal, Resumes and application data.

4.   To Delete a previously applied resume -
By clicking on the  icon then click Delete.

5.   To Upload a new resume-
click on Upload resume


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