
Concept explainers
“SELECT” command:
The “SELECT” command is used to retrieve data in a
Syntax for selecting values from the table is as follows:
SELECT STUDENT_ID FROM STUDENT;
- The given query is used to display each student ID from “STUDENT” table.
“COUNT” function:
- It is the one function of aggregate function.
- The “COUNT” function is used to compute the number of rows in a table.
Example:
The example for “COUNT” function is given below:
SELECT COUNT(*) FROM STUDENT WHERE MARK_CREDIT >= 90;
The above query is used to display the number of students whose mark credit is greater or equal to “90” by using “COUNT” function.
- From the given query, the asterisk (*) represent any column.
- User can also count the number of rows in a query by selecting a particular column instead of using the asterisk.
- The below example is as follows
SELECT COUNT(STUDENT_ID) FROM STUDENT WHERE MARK_CREDIT >= 90;
- The below example is as follows
“ORDER BY” Clause:
- User can sort the data in specific order using “ORDER BY” clause.
- The column on which to sort data is called a sort key or a simple key.
- To sort the output, use an “ORDER BY” clause followed by the sort key.
- If the user does not indicate a sort order, the output displayed in default order that is ascending order.
Example:
The example for “ORDER BY” clause is given below:
SELECT STUDENT_ID, STUDENT_NAME, STUDENT_CREDIT FROM STUDENT ORDER BY STUDENT_CREDIT;
The above query is used to list student ID, name and credit for each student with ascending order of student credit using an “ORDER BY” clause.
- From the given query, the sort key is “STUDENT_CREDIT”. So, the rows are sorted in ascending order by “STUDENT_CREDIT”.
“GROUP BY” Clause:
- User can group the data using “GROUP BY” clause.
- This clause allows the user to group data on a specific column and then computes statistics when user preferred.
Example:
The example for “GROUP BY” clause is given below:
SELECT CUSTOMER_NAME, SUM(AMOUNT) FROM CUSTOMERS GROUP BY CUSTOMER_NAME;
The above query is used to list the customer name and the sum of amount using “GROUP BY” clause.

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Chapter 4 Solutions
A Guide to SQL
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- It is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardIt is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardIt is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forward
- It is possible to sort an array of n values using pipeline of n filter processes.The first process inputs all the values one at a time, keep the minimum, and passes the others on to the next process. Each filter does the same thing; it receives a stream of values from the previous process, keep the smallest, and passes the others to the next process. Assume each process has local storage for only two values--- the next input value and the minimum it has seen so far. (a) Developcode for filter processes. Declare the channels and use asynchronous message passing. Hint:Define an array of channels value[n] (int), and a set of filter processes Filter[i = 0 ton-1]. Each process Filter[i] (where 0 <= i <= n-2) receives a stream of integers through channelvalue[i], keeps the smallest, and sends all other integers to channel value[i+1]. The last processFilter[n-1] receives only one integer through channel value[n-1] and does not need to send anyinteger further.arrow_forwardI need help: Challenge: Assume that the assigned network addresses are correct. Can you deduce (guess) what the network subnet masks are? Explain while providing subnet mask bits for each subnet mask. [Hint: Look at the addresses in binary and consider the host ids]arrow_forwardI would like to know if my answer statment is correct? My answer: The main difference is how routes are created and maintained across different networks. Static routing establishes router connections to different networks from the far left and far right. The Dynamic routing focus emphasizes immediate connection within the router while ignoring the other connections from different networks. Furthermore, the static routing uses the subnet mask to define networks such as 25.0.0.0/8, 129.60.0.0/16, and 200.100.10.0/30, which correspond to 255.0.0.0, 255.255.0.0, and 255.255.255.252. On the other hand, dynamic routing uses the wildcard mask to inverse the subnet mask, where network bits become 0 and host bits become 1, giving us 0.0.0.255, 0.0.255.255, and 0.0.0.3. Most importantly, the CLI commands used for Static and Dynamic routing are also different. For static routing, the “IP route” corresponds with the network, subnet mask, and next-hop IP address. In contrast, dynamic routing uses…arrow_forward
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