Write the following queries in SQL: Display a list of all instructors, showing their ID, name, and the number of sections that they have taught. Make sure to show the number of sections as 0 for instructors who have not taught any section. Your query should use an outer join, and should not use scalar subqueries. By using the university schema provided by db-book.com the following queries were done on the university database. The first query uses an outer join which works similar to the join
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records which combines each row from the first table with each row of the second table. Cross JOIN Syntax is, SELECT column-name-list from table-name1 CROSS JOIN table-name2; Example of Cross JOIN The class table, ID | NAME | 1 | abhi | 2 | adam | 4 | alex | The class_info table, ID | Address | 1 | DELHI | 2 | MUMBAI | 3 | CHENNAI | Cross JOIN query will be, SELECT * from class
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into the cursor each tuple of the relation, we can write a program to read and process the value of each such tuple. If the relation is stored, we can also update or delete the tuple at the current cursor position. The example below illustrates a cursor loop. It uses our example relation T1(e,f) whose tuples are pairs of integers. The program will delete every tuple whose first component is less than the second, and insert the reverse tuple into T1. 1) DECLARE /* Output variables
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Thank you Ptlm Oliver! I’m excited to hear if you receive the grant. Please let me know if there is anything that the Woonsocket Health Equity Zone (HEZ) collaborative can do to support any pedestrian safety initiative or funding opportunity that the WPD is doing. Come either in Sept/Oct the collaborative will be launching the Healthy Environments workgroup in Woonsocket, which has a primary focus on pedestrian safety and walkability. I would love to schedule a time with you, Patrick and if possible
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Week 4 Melissa Maxwell Assignment Chapter 8 1. CREATE TABLE CUSTOMER CUST_NUM NUMBER PRIMARY KEY, CUST_LNAME VARCHAR CUST_FNAME VARCHAR CUST_BALANCE NUMBER; CREATE TABLE CUSTOMER_2 CUST_NUM NUMBER PRIMARY KEY, CUST_LNAME VARCHAR, CUST_FNAME VARCHAR CREATE TABLE INVOICE INV_NUM NUMBER PRIMARY KEY, CUST_NUM NUMBER, INV_DATE DATE, INV_AMOUNT NUMBER); 2. INSERT INTO CUSTOMER VALUES (1000, ‘SMITH’, JEANNE’ 11050.11); INSERT INTO CUSTOMER VALUES(1001, ‘ORTEGA’, ‘JUAN’ 840.92); INSERT INTO CUSTOMER_2
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only work properly if relations are Union-Compatible, which is based on the names of the relation attributes that must be the same and their data types must be alike. Being compatible does not mean the data types have to be exactly the same. For example, both data types can be used to store numeric values such as NUMBER and SMALLINT as well as character (string) values such as VARCHAR and CHAR. An SQL IN statement can be used with where clause to list a set of matching records of a table. We can
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*****Library for OleDb command and Connection Imports System.Data.OleDb ** *****Declaration for command, connection and Reader Public dcom As OleDbCommand Public dcon As OleDbConnection Public dbrd As OleDbDataReader Public sql As String ** ******Connection String *******2000 – 2003 Version (.mdb format) **A. dcon = New OleDbConnection("provider=microsoft.jet.oledb.4.0;data source = " & Application.StartupPath & "\maindata.mdb") dcon.Open() **B. dcon = New OleDbConnection("provider=microsoft
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Public Class RBTree Public root As RBTNode Dim null As RBTNode Public Sub New(ByVal data As Integer) null = New RBTNode("-1") null.color = "black" root = New RBTNode(data) root.color = "black" root.Left = null root.Left.parent = root root.right = null root.right.parent = root End Sub Public Sub New() End Sub Public Sub Insert(ByVal data As Integer) null = New RBTNode("-1")
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| Database systems | Assignment # 1 | Areeba kamil (34421) | BSCS 4A | 10/4/2015 | 4.7. Consider the LIBRARY relational database schema shown in Figure 4.6. Choose the appropriate action (reject,cascade,set to NULL,set to default) for each referential integrity constraint, both for the deletion of a referenced tuple and for the update of a primary key attribute value in a referenced tuple.Justify your choices. CREATE DATABASE library; CREATE TABLE book( book_id int, title varchar(50)
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SELECT column_list FROM table-name [WHERE Clause] [GROUP BY clause] [HAVING clause] [ORDER BY clause]; SELECT name, salary, salary*1.2 AS new_salary FROM employee WHERE new_salary /salary*1.2 > 30000; SELECT first_name, last_name, subject FROM student_details WHERE subject IN/NOT IN ('Maths', 'Science'); Select * from product p where EXISTS (select * from order_items o where o.product_id = p.product_id) -------------------------------------------------
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