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What is the difference between RDBMS and DBMS?
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Answer Posted / saurav majumder
DBMS creates Data and Meta Data
RDBMS creates relations between tables and rows.
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Hi, I am new in oracle(SQL), could anyone help me in writing a correct SQL. Below is the table structure. Table: Subsc Fields: 1. Sub_no (this field will hold values of subscriber nos, for e.g. S111111, S222222, S333333, S444444, etc.) 2. s_status (this field will hold values for different status of subscriber, for e.g. 'A', 'S', 'C', etc.) 3. cus_id (this field will hold values of bill nos for e.g. 11111111, 22222222, 33333333, 44444444, etc.) Table: Bill Fields: 1. Bill_no this field will hold values of bill nos for e.g. 11111111, 22222222, 33333333, 44444444, etc.) 2. b_status = (this field will hold values for different status of bill for e.g. 'O', 'C', 'S', etc.) Note: 1. The Sub_no is a Primary key of Subsc table. 2. The cus_id is a foreign in Subsc table (referred from Bill_no field of Bill table) 3. The Bill_no field is the Primary key of Bill table. Query A --> I wrote a query to select cus_id/Bill_no which is in status open (b_status = 'O') and having more than two active subscriber (i.e. S_status = 'A') in it ( i.e. more the two subscribers in same bill). select s.cus_id from subsc s where exists (select 1 from bill where bill_no = s.cus_id and b_status = 'O') and s_status = 'A' group by s.cus_id having count(sub_no) = 2 Problem : The above query will give the cus_id (or rather bill_no) which are in open status (b_status ='O) and which are having TWO ACTIVE Subscribers (s_status ='A') in it. However, this query will also lists the cus_id/bill_no which are having more than TWO subscribers in it (but only two subscriber will be in Active status (s_status = 'A') and the others will be in s_status = 'C' or s_status = 'S'. Help needed: I want to write a query which will fetch ONLY the cus_id/bill_no which are in open status (b_status ='O') and which are having ONLY TWO ACTIVE subscribers (s_status ='A') in it. B--> If I include the sub_no in the above query then NO row are returned. select s.cus_id, s.sub_no from subsc s where exists (select 1 from bill where bill_no = s.cus_id and b_status = 'O') and s_status = 'A' group by s.cus_id, s.sub_no having count(sub_no) = 2 Help needed: I want to modify the above query which will fetch ONLY the cus_id/bill_no which are in open status (b_status ='O') and which are having ONLY TWO ACTIVE subscribers (s_status ='A') in it ALONG with the sub_no. Thanks a lot in advance. Regards, Nitin
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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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