DB2 use the multiple indexes to satisfy multiple predicates in a SELECT statement that are joined by an AND or OR.

SELECT * FROM table_name ORDER BY column_name desc Fetch first 5 rows only;

To insert a NULL, move -1 to the null indicator

To insert a valid value, move 0 to the null indicator

1. Based on the lock-size given while creating the table space

2. Programer can direct the DB2 what lock-size to use

3. If lock - size ANY is specified, DB2 usually chooses a lock-size of PAGE

Cursor is a programming device that allows the SELECT to find a set of rows but return them one at a time. Cursor should be used because the host language  can deal with only one row at a time.

Use SELECT DISTINCT ..... in DB2 query

The DISTINCT keyword can be used to return only distinct (different) values.

Syntax:

     SELECT DISTINCT column1,column2

      FROM table_name;

public class Test {

    public void replaseSpaces(char[] str, int length) {

        int spaceCount = 0, newLength = 0, i = 0;

        for(i = 0; i < length; i++) {

            if (str[i] == ' ') 

                spaceCount++;

        }

        newLength = length + (spaceCount * 2);

        str[newLength] = '\0';

        for(i = length - 1; i >= 0; i--) {

            if (str[i] == ' ') {

                str[newLength - 1] = '0';

                str[newLength - 2] = '2';

                str[newLength - 3] = '%';

                newLength = newLength - 3;

            }

            else {

                str[newLength - 1] = str[i];

                newLength = newLength - 1;

            }

        }

        System.out.println(str);

    }

    public static void main(String[] args) {

        Test tst = new Test();

        char[] ch = {'t', 'h', 'e', ' ', 'd', 'o', 'g', ' ', ' ', ' ', ' ', ' ', ' '};

        int length = 6;

        tst.replaseSpaces(ch, length);  

    }

}

This problem isn't too hard. You just need to start at the head of the list, and iterate to the end. At each node, swap the values of pNext and pPrev. Finally, set pHead to the last node in the list.

Node * pCurrent = pHead, *pTemp;

while (pCurrent)

{

  pTemp = pCurrent->pNext;

  pCurrent->pNext = pCurrent->pPrev;

  pCurrent->pPrev = temp;  

  pHead = pCurrent;

  pCurrent = temp;

}