Write a query to find the name of employees those who have
joined on Monday.(based on column hire_date)
Answer Posted / sankarapandian
select *,upper(datename(dw,gedate())) as dayofweek from
employee where upper(Convert(varchar(15),datename dw,gedate
())))='MONDAY'
Is This Answer Correct ? | 1 Yes | 4 No |
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Table1: Col1 col2 1 2 10 3 4 89 5 6 Table:2 Col1 col2 3 2 9 5 4 7 6 87 With the help of table1 and table2 write a query to simulate the fallowing results. Output1: Col1 col2 1 2 2 3 3 4 4 5 5 6 Output2: Col1 col2 2 3 10 4 5 89 6 7 1.Write query for single row to multiple row using sql statements. Eg:a,b,c,d,e,f Change to A B C D E F 2. Write query for multiple row to single row using sql statements. Eg2 A B C D E F Change to Eg:a,b,c,d,e,f Table1: Col1 col2 8 5 2 9 4 2 5 1.Write a query to select all the rows from a table1,if the value of A is null then corresponding B’s value should be printed in A’s value.if the value of A is null in that table then corresponding B’s value should be printed as 30. 2. write a query to find the sum of A and B .display the max among both. 3.write a query to find total number of rows from table 1. Note: if any column value is null in a row then that row should be considered as 2 rows. 4.write a query to display all the records of table1 except A containg 2 as well B containg 5. 5.rewrite the fallowing without using join and group by. Select b.title,max(bc.returneddate –bc.checkoutdate)” mostdaysout” From bookshelf_checkout bc, Book shelf B Where bc.title(+)=b.title Group by b.title. 6.rewrite fallowing query Select id_category from category_master X where exists (select 1 from sub_category Y where X.id_category=Y.id_category) Customer: Name phone1 phone2 phone3 bitwise A 23456 67890 12345 --- B 67459 89760 37689 --- Don’t_call Col1 67890 37689 1.q) update the customer table of bitwise with 1 or 0. Exists in don’t_call table menas show -1 Other wise -0. Output. Name bitwise A 010 B 010
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