@@Manikandan-pq9dm the second result is correct bro, in the first result you can use count(employee_id) if you want ,else it throws an error. select avg(salary),count(*), dept from employee group by dept having avg(salary) >55000 or count(*) >2 This works well and good for me. You can't use " employee " since it is the table name.
Thank you for ur python course, it's use full form me 🙏🙏🙏Hi bro plz conduct full stack developer course it will be most helpful for me becoz ur teaching is easy to understand,
last question's answer: select department from employee group by department having avg(salary)>55000 || count(department)>2 Hope you all get this. Or else correct it if I'm wrong🙂
I guess he has provided two requirements. Here they are, Select departments ( select avg(salary), departments from group by departments) from ; Select departments ( select count(*) as no_of_employees, departments from group by departments having no_of_employees >2) from ;
@@jjbala2198 because they are haing more than 2 employees..Instant of using this you can try this bro:select avg(salary),count(department) from employee group by department having avg(salary)>55000 or count(*)>=2;
Anna, when my mam taught me at school I did not get anything but you made me understand having clause in just 5 mins. Thank you so much😇 but what is the answer for this question 4:43 it's driving me mad with out the answer. I solved the first half of the question but I can't even understand the last part of the question. select avg(salary),dept from employee group by dept having avg(salary)>55000 [ I hope this is the answer for the first half🤓]
Last Question Answer :- select avg(salary) , count(*), department from employee group by department having avg(salary) > 55000.0000 or count(*) > 2; Output: avg(salary) count(*) department 61000.0000 2 IT 58500.0000 2 FINANCE Reason: IT and Finance Department Has Average Salary more than 55,000 , So it is displayed. What about Departments with more than 2 Employees!!, Every Departments in the Data has only 2 Employees, So Even if you omit this Condition "count(*) > 2", The Output will be Same becoz there is No department with more than 2 Employees.
Why was the answer to the last question not provided? Are you trying to use your strategy to encourage joining your paid class? If the answer to the last question was provided, it would be useful to us.
select avg(salary),department from employee group by department having avg(salary)>55000 limit 2
*
select avg(salary),count(*),department from employee
group by department
having avg(salary)>55000 or count(*)>2 ;*
count(employee) thana varum?
@@Manikandan-pq9dm question la count of employees more than 2 nu dha iruk so employees dha varum ig
@@Manikandan-pq9dm the second result is correct bro, in the first result you can use count(employee_id) if you want ,else it throws an error.
select avg(salary),count(*), dept from employee group by dept
having avg(salary) >55000 or
count(*) >2
This works well and good for me.
You can't use " employee " since it is the table name.
join and subquery pathi explain Pannunga bro waiting so far
Thank you for ur python course, it's use full form me 🙏🙏🙏Hi bro plz conduct full stack developer course it will be most helpful for me becoz ur teaching is easy to understand,
You are the best teacher
Select department from employees group by department having avg(salary) > 55000 or count(employee_id) > 2;
last question's answer:
select department from employee group by department having avg(salary)>55000 || count(department)>2
Hope you all get this. Or else correct it if I'm wrong🙂
I guess he has provided two requirements. Here they are,
Select departments ( select avg(salary), departments from group by departments) from ;
Select departments ( select count(*) as no_of_employees, departments from group by departments having no_of_employees >2) from ;
@@devanathanelangovan4961 select avg(salary),count(department) from employee group by department having avg(salary)>55000 or count(*)>=2;
SELECT avg(salary), department from employee
group by department
having avg(salary)>$55000
limit 10;
Bro can you explain why should we use limit 10 😢😢
I tried without limit it's actually worked
@@jjbala2198 because they are haing more than 2 employees..Instant of using this you can try this bro:select avg(salary),count(department) from employee group by department having avg(salary)>55000 or count(*)>=2;
Please upload remaining videos of SQL soon bro. Waiting for so long🥹
Bro Html And Css Video Detail ah Pottu irrukinga Learn panna Use full ah irruku So Apdye Java Script Konjam Detail ah Video podunga Bro ❤
select avg(salary),department from employee group by department order by avg(salary) asc limit 2
Select avg(sal),count(*)department from employees group by department having avg (salary)>55000 and count(*) > 2;
select avg(salary),count(*), dept from employee group by dept
having avg(salary) >55000 or
count(*) >2
Count(*)>2 instead of use limit 2
Select avg(salary), department from employee group by department having avg(salary) > 55000
Result grid
61000,58500
Bro foreign key pathi podunga pls
Bro SQL la Ella topics um cover pannunga innum neraiya topics irukku
Bro DCL, VIEWS enu db objects lam solitharalaye ana playlist end airuchi😭Humble req to finish all the topics involved.
Select departments,avg(salary) from employee group by department having salary>55000 or count(department)>2
Please upload Java script playlist brother.
Your way of teaching is Clear.
Anna, when my mam taught me at school I did not get anything but you made me understand having clause in just 5 mins. Thank you so much😇
but what is the answer for this question 4:43 it's driving me mad with out the answer. I solved the first half of the question but I can't even understand the last part of the question.
select avg(salary),dept from employee group by dept having avg(salary)>55000
[ I hope this is the answer for the first half🤓]
select avg(salary),count(department) from employee group by department having avg(salary)>55000 or count(*)>=2;
select avg(salary) from employee group by department order by avg(salary) asc limit =2
Answer for the question is select avg( salary), department from employee group by department having avg(salary)>55000 limit2
wrong
select avg(salary) as avg_salary, department from employee group by department having avg_salary
great.!
Bro JavaScript video podunga pls
Select avg(sal) from employees
Group by department
Having avg(sal) < 55,000 or count (*)>2
select avg(salary),department from employee group by department having avg(salary)>55000 is the answer
Javascript course podunga bro
SELECT AVG(salary), department FROM employee
GROUP BY department
HAVING AVG(salary)> 55,000
OFFSET 2;
Power Bi videos podunga bro
Waiting for next part bro
bro completed is this correct
select dept,avg(salary),count(dept) from employee group by dept having avg(salary) > 55000 or count(dept) > 2
Pls upload javascript workshop video
SELECT
AVG(salary), COUNT(*), department
FROM
employee
GROUP BY department
HAVING AVG(salary) > 55000 OR COUNT(*) < 2;
Select Count(EmployeeID),Department From Employee Group by Department Having avg(Salary)>55000 limit 2
Wrong bro
@@MasterJD-pz4lz ohh ok if you don't mind please explain.nanu therinchu kura
Anna nenga irukarathu nalla yenaku romba romba easya html css js kathukitta romba romba thanks
And finally last request anna c & c++
Video poduvigala
Super bro
After finishing thiss we want javascript bruhhhhhhh😢
Thanks bro put next video asap
Last Question Answer :-
select avg(salary) , count(*), department
from employee
group by department
having avg(salary) > 55000.0000 or count(*) > 2;
Output:
avg(salary) count(*) department
61000.0000 2 IT
58500.0000 2 FINANCE
Reason:
IT and Finance Department Has Average Salary more than 55,000 , So it is displayed.
What about Departments with more than 2 Employees!!,
Every Departments in the Data has only 2 Employees, So Even if you omit this Condition "count(*) > 2", The Output will be Same becoz there is No department with more than 2 Employees.
Bro fan request, pls upload C programming language 😢
SELECT count(*) AS No_employees, department FROM EMPLOYEE GROUP BY DEPARTMENT HAVING No_employees>2
🎉 first comment
Can you please explain Triggers ?
It would be very useful for me !
Bro nenga SQL Join pathi full video podunga bro
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And clear.then how I can do it in mobile phone
Broo RPA pathi solunga
Django framework using python sollithanga
Bro school students LinkedIn use pannalama bro school students eppadi use pandra dhu bro ........
Java script upload pannunga bro...
pls upload remaining topics in sql bro
Bro..how to print prime numbers less than or equal to 1000 ..in mySQL intha ques CLR panu anna
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R Programming podunga bro
Upload javascript
👍
do people use Version Control Software for SQL code?
how do people keep version control of SQL code?
answer for last question?
Anna Networking Class edunga naa
sir video call application js how to create
Bro what about join command bro
Bro laptop illa enna bro pandrathu but I was study the theorically
Bro..pls put Java script video bro
Bro c language poduga
Bro python ku pota mari varaha edachu oru language full ha poduga bro
bro pls ipdi zoom panathinga pathutae irukura appo kannu valikuthu
Integrating SQL in a PL/SQL program la integrating enna bro.????
Why was the answer to the last question not provided? Are you trying to use your strategy to encourage joining your paid class? If the answer to the last question was provided, it would be useful to us.
loose punda
Hi bro pls do. Net
why there is no JS video in your video list
Hii
Hacking online course erutha soluga bro
Javascript video podunga bro plssss😢😢
Broooii
C# please😵
bro why courses not available ?
What is IT work? Anga yanna pannaum sollunga proo plss
Coding !
Select avg(salary),count(*) department from employee group by department having avg( salary )>55000 or count(*)2
select avg(salary),department from employee group by department having avg(salary)>55000 limit 2
Select avg( salary),department from employee group by department having avg(salary) >55000