How to Add or Subtract Days, Months, or Years from a Date in C#

Manipulating dates is a common task in C# applications, whether for scheduling, logging, or calculations.

The DateTime and DateOnly structures provide built-in methods to add or subtract days, months, years, hours, and minutes efficiently.

Adding and Subtracting Days

Use the AddDays method to modify a DateTime instance:

using System;

class Program
{
    static void Main()
    {
        DateTime today = DateTime.Now;
        DateTime nextWeek = today.AddDays(7);
        DateTime lastWeek = today.AddDays(-7);

        Console.WriteLine($"Today: {today:yyyy-MM-dd HH:mm}");
        Console.WriteLine($"Next Week: {nextWeek:yyyy-MM-dd HH:mm}");
        Console.WriteLine($"Last Week: {lastWeek:yyyy-MM-dd HH:mm}");
    }
}

Adding and Subtracting Months

Use the AddMonths method to adjust the month while automatically handling month-end variations:

DateTime currentDate = new DateTime(2025, 3, 31);
DateTime nextMonth = currentDate.AddMonths(1);
DateTime previousMonth = currentDate.AddMonths(-1);

Console.WriteLine($"Current Date: {currentDate:yyyy-MM-dd HH:mm}");
Console.WriteLine($"Next Month: {nextMonth:yyyy-MM-dd HH:mm}");
Console.WriteLine($"Previous Month: {previousMonth:yyyy-MM-dd HH:mm}");

Adding and Subtracting Years

Use the AddYears method to adjust the year, handling leap years automatically:

DateTime date = new DateTime(2024, 2, 29);
DateTime nextYear = date.AddYears(1);
DateTime previousYear = date.AddYears(-1);

Console.WriteLine($"Original Date: {date:yyyy-MM-dd HH:mm}");
Console.WriteLine($"Next Year: {nextYear:yyyy-MM-dd HH:mm}");
Console.WriteLine($"Previous Year: {previousYear:yyyy-MM-dd HH:mm}");

Adding and Subtracting Hours

Use the AddHours method to modify the hour component:

DateTime now = DateTime.Now;
DateTime inFiveHours = now.AddHours(5);
DateTime fiveHoursAgo = now.AddHours(-5);

Console.WriteLine($"Current Time: {now:yyyy-MM-dd HH:mm}");
Console.WriteLine($"In 5 Hours: {inFiveHours:yyyy-MM-dd HH:mm}");
Console.WriteLine($"5 Hours Ago: {fiveHoursAgo:yyyy-MM-dd HH:mm}");

Adding and Subtracting Minutes

Use the AddMinutes method to modify the minute component:

DateTime currentTime = DateTime.Now;
DateTime inThirtyMinutes = currentTime.AddMinutes(30);
DateTime thirtyMinutesAgo = currentTime.AddMinutes(-30);

Console.WriteLine($"Current Time: {currentTime:yyyy-MM-dd HH:mm}");
Console.WriteLine($"In 30 Minutes: {inThirtyMinutes:yyyy-MM-dd HH:mm}");
Console.WriteLine($"30 Minutes Ago: {thirtyMinutesAgo:yyyy-MM-dd HH:mm}");

Using DateOnly for Simpler Date Manipulation

For applications that don't require time components, DateOnly (introduced in .NET 6) provides a cleaner approach:

DateOnly today = DateOnly.FromDateTime(DateTime.Now);
DateOnly futureDate = today.AddDays(30);

Console.WriteLine($"Today: {today}");
Console.WriteLine($"30 Days Later: {futureDate}");

Conclusion

C# provides built-in methods for adjusting dates effortlessly. Whether working with DateTime or DateOnly, these functions ensure accurate date calculations, even when dealing with leap years, month-end scenarios, hours, and minutes.

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Related

When working with SQL Server, you may often need to count the number of unique values in a specific column. This is useful for analyzing data, detecting duplicates, and understanding dataset distributions.

Using COUNT(DISTINCT column_name)

To count the number of unique values in a column, SQL Server provides the COUNT(DISTINCT column_name) function. Here’s a simple example:

SELECT COUNT(DISTINCT column_name) AS distinct_count
FROM table_name;

This query will return the number of unique values in column_name.

Counting Distinct Values Across Multiple Columns

If you need to count distinct combinations of multiple columns, you can use a subquery:

SELECT COUNT(*) AS distinct_count
FROM (SELECT DISTINCT column1, column2 FROM table_name) AS subquery;

This approach ensures that only unique pairs of column1 and column2 are counted.

Why Use COUNT DISTINCT?

  • Helps in identifying unique entries in a dataset.
  • Useful for reporting and analytics.
  • Efficient way to check for duplicates.

By leveraging COUNT(DISTINCT column_name), you can efficiently analyze your database and extract meaningful insights. Happy querying!

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In C#, you can format an integer with commas (thousands separator) using ToString with a format specifier.

int number = 1234567;
string formattedNumber = number.ToString("N0"); // "1,234,567"
Console.WriteLine(formattedNumber);

Explanation:

"N0": The "N" format specifier stands for Number, and "0" means no decimal places. The output depends on the culture settings, so in regions where , is the decimal separator, you might get 1.234.567.

Alternative:

You can also specify culture explicitly if you need a specific format:

using System.Globalization;

int number = 1234567;
string formattedNumber = number.ToString("N0", CultureInfo.InvariantCulture);
Console.WriteLine(formattedNumber); // "1,234,567"
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Storing passwords as plain text is dangerous. Instead, you should hash them using a strong, slow hashing algorithm like BCrypt, which includes built-in salting and resistance to brute-force attacks.

Step 1: Install BCrypt NuGet Package

Before using BCrypt, install the BCrypt.Net-Next package:

dotnet add package BCrypt.Net-Next

or via NuGet Package Manager:

Install-Package BCrypt.Net-Next

Step 2: Hash a Password

Use BCrypt.HashPassword() to securely hash a password before storing it:

using BCrypt.Net;

string password = "mySecurePassword123";
string hashedPassword = BCrypt.HashPassword(password);

Console.WriteLine(hashedPassword); // Output: $2a$12$...

Step 3: Verify a Password

To check a user's login attempt, use BCrypt.Verify():

bool isMatch = BCrypt.Verify("mySecurePassword123", hashedPassword);
Console.WriteLine(isMatch); // Output: True

Ensuring proper hashing should be at the top of your list when it comes to building authentication systems.

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