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artcher [175]
3 years ago
8

Q. Which protocol would best serve to authorize users to access directory services?

Computers and Technology
1 answer:
Arlecino [84]3 years ago
6 0

Answer:

Lightweight directory access protocol (LDAP)

Explanation:

Lightweight directory access protocol is a protocol that provides a common open protocol for interfacing and querying a directory service information given by a network operating system. LDAP is an application layer protocol that uses port number 389 via tcp and udp. LDAP queries are transmitted in text and allows for all or some data be queried anonymously.

Lightweight directory access protocol, ordinarily does not require user authentication connection, but this can be configured to ask for user credentials.

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Your computer is configured to obtain an ipv4 address and dns server address automatically. what utility will help you to find t
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3 years ago
You should process the tokens by taking the first letter of every fifth word,starting with the first word in the file. Convert t
zhenek [66]

Answer:

See explaination

Explanation:

import java.io.File;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class SecretMessage {

public static void main(String[] args)throws IOException

{

File file = new File("secret.txt");

StringBuilder stringBuilder = new StringBuilder();

String str; char ch; int numberOfTokens = 1; // Changed the count to 1 as we already consider first workd as 1

if(file.exists())

{

Scanner inFile = new Scanner(file);

StringTokenizer line = new StringTokenizer(inFile.nextLine()); // Since the secret.txt file has only one line we dont need to loop through the file

ch = line.nextToken().toUpperCase().charAt(0); // Storing the first character of first word to string builder as mentioned in problem

stringBuilder = stringBuilder.append(ch);

while(line.hasMoreTokens()) { // Looping through each token of line read using Scanner.

str= line.nextToken();

numberOfTokens += 1; // Incrementing the numberOfTokens by one.

if(numberOfTokens == 5) { // Checking if it is the fifth word

ch = str.toUpperCase().charAt(0);

stringBuilder = stringBuilder.append(ch);

numberOfTokens =0;

}

}

System.out.println("----Secret Message----"+ stringBuilder);

}

}

}

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The answer is 
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bezimeni [28]
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3 years ago
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