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NEW QUESTION: 1
Which of the following statements about the Lookup Stage is correct?
A. The default partition method for Lookup is Hash
B. It combines data horizontally
C. It is a Real-Time Stage
D. It can have only one input link defined
Answer: B
NEW QUESTION: 2
Which three statements about subqueries are true? (Choose three)
A. A single row subquery can use the IN operator.
B. A single row subquery can retrieve only one row but many columns.
C. A multiple row subquery can use the "=" operator.
D. A multiple row subquery can be compared by using the ">" operator.
E. A multiple row subquery can retrieve multiple rows and multiple columns.
F. A single row subquery can retrieve only one column and one row.
Answer: A,B,E
Explanation:
B: A single row sub-query can retrieve only one row but many columns.
C: A multiple row sub-query can retrieve one row or multiple rows and multiple columns.
E: A single row subquery can use the "IN" Operator. Although the IN Operator is a multiple row operator it can be used for comparisons with single-row subqueries and it will not result in an error
Incorrect Answers
A: A single row sub-query can retrieve only one row, but many columns..
D: Multiple row subquery cannot use the ">" Operator because this is a single row operator.
F: A multiple row sub-query cannot use the "=" operator.
OCP Introduction to Oracle 9i: SQL Exam Guide, Jason Couchman, p. 150-165 Chapter 4: Subqueries
NEW QUESTION: 3
Given:
import java.io.IOException;
import java.io.file.Path;
import java.io.file.Paths;
public class Path12 {
public static void main(String s[]) throws IOException { Path path = Paths.get("\\sales\\quarter\\..\\qtrlreport.txt");
path.relativize(Paths.get("\\sales\\annualreport.txt"));
if(path.endsWith("annualreport.txt")) {
System.out.println(true);
} else {
System.out.println(false);
}
System.out.println(path);
}
}
What is the result?
A. false \sales\quarter\ . . \qtrlreport.txt
B. false \quarter\ . . \qtrlreport.txt
C. true . . \ . . \ . . \ annualreport.txt
D. true \ . . \ . . \annualreport.txt
Answer: A
Explanation:
The relativize method that can be used to construct a relative path between two
paths.
relativize
Path relativize(Path other)
Constructs a relative path between this path and a given path.
Parameters:
other - the path to relativize against this path
Returns:
the resulting relative path, or an empty path if both paths are equal
Note:
Relativization is the inverse of resolution. This method attempts to construct a relative path that
when resolved against this path, yields a path that locates the same file as the given path. For
example, on UNIX, if this path is "/a/b" and the given path is "/a/b/c/d" then the resulting relative
path would be "c/d". Where this path and the given path do not have a root component, then a
relative path can be constructed. A relative path cannot be constructed if only one of the paths
have a root component. Where both paths have a root component then it is implementation
dependent if a relative path can be constructed. If this path and the given path are equal then an
empty path is returned.
For any two normalized paths p and q, where q does not have a root component,
p.relativize(p.resolve(q)).equals(q)
When symbolic links are supported, then whether the resulting path, when resolved against this
path, yields a path that can be used to locate the same file as other is implementation dependent.
For example, if this path is "/a/b" and the given path is "/a/x" then the resulting relative path may
be "../x". If "b" is a symbolic link then is implementation dependent if "a/b/../x" would locate the
same file as "/a/x".
