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NEW QUESTION: 1
Which of the following statements about the Lookup Stage is correct?
A. It can have only one input link defined
B. It combines data horizontally
C. The default partition method for Lookup is Hash
D. It is a Real-Time Stage
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 column and one row.
C. A single row subquery can retrieve only one row but many columns.
D. A multiple row subquery can retrieve multiple rows and multiple columns.
E. A multiple row subquery can use the "=" operator.
F. A multiple row subquery can be compared by using the ">" operator.
Answer: A,C,D
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. true \ . . \ . . \annualreport.txt
C. true . . \ . . \ . . \ annualreport.txt
D. false \quarter\ . . \qtrlreport.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".
