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Mice21 [21]
3 years ago
14

Your Web page provides a form for users to add their names to a membership list, stored in a text file. You are writing a progra

m to process the form submission. How should your program open the file? Read operations Write operations Append operations This will not work: the form will be used by different people at different times.
Computers and Technology
1 answer:
vlada-n [284]3 years ago
3 0

Answer:

For the above condition, and as mentioned none of the mode will work, and we need to make use of the JSON format. And we nee to use the JavaScript, and we don't need to mention the mode in JavaScript. Thus, this is nowadays used solution for this old problem that existed when we used to code using c++ and c.

Explanation:

You need to write over this text file, and you also need to read. Hence, you need to open the file with the three privileges, and the append will be required,and this way we can only move line by line. However, this is an old style. Now we save in text file in JSON format. The data collected  from Form is converted to JSON format and written in text file. And as required data can be read or written, and as required.  So if we use only JSON format, then we need not mention the mode, as its not required these days in new languages like JavaScript.

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Answer:

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Explanation:

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3 years ago
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You have a notebook computer and wish to connect to an IEEE 802.11ac wireless network. The computer does not have a built-in WLA
BARSIC [14]

Answer:

Wireless USB 2.0

Explanation:

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3 years ago
Write a JavaScript program to generate the following pattern but the number of rows should be user input.
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Answer:

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3 years ago
Write a c++ program to print even numbers from 1 to 20​
Novay_Z [31]

Answer:

#include <bits/stdc++.h>

using namespace std;

// Function to print even numbers

void printEvenNumbers(int N)

{

cout << "Even: ";

for (int i = 1; i <= 2 * N; i++) {

// Numbers that are divisible by 2

if (i % 2 == 0)

cout << i << " ";

}

}

// Function to print odd numbers

void printOddNumbers(int N)

{

cout << "\nOdd: ";

for (int i = 1; i <= 2 * N; i++) {

// Numbers that are not divisible by 2

if (i % 2 != 0)

cout << i << " ";

}

}

// Driver code

int main()

{

int N = 20;

printEvenNumbers(N);

printOddNumbers(N);

return 0;

}

Explanation:

Note: This will find both odd and even numbers, you have to change the number above to the number of your choice

For even numbers

Even number are numbers that are divisible by 2.

To print even numbers from 1 to N, traverse each number from 1.

Check if these numbers are divisible by 2.

If true, print that number.

For odd numbers

Odd number are numbers that are not divisible by 2.

To print Odd numbers from 1 to N, traverse each number from 1.

Check if these numbers are not divisible by 2.

If true, print that number

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3 years ago
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The number and letter are switched. This question needs more context to get a accurate answer.
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