String interpolation, introduced in C# 6.0, provides a more readable and concise way to format strings compared to traditional concatenation (+) or string.Format(). Instead of manually inserting variables or placeholders, you can use the $ symbol before a string to directly embed expressions inside brackets.
string name = "Walt"; string job = 'Software Engineer'; string message = $"Hello, my name is {name} and I am a {job}"; Console.WriteLine(message);
This would produce the final output of:
Hello, my name is Walt and I am a Software Engineer
String interpolation can also be chained together into a multiline string (@) for even cleaner more concise results:
string name = "Walt"; string html = $@" <div> <h1>Welcome, {name}!</h1> </div>";
When working with URLs in C#, encoding is essential to ensure that special characters (like spaces, ?, &, and =) don’t break the URL structure. The recommended way to encode a string for a URL is by using Uri.EscapeDataString(), which converts unsafe characters into their percent-encoded equivalents.
string rawText = "hello world!"; string encodedText = Uri.EscapeDataString(rawText); Console.WriteLine(encodedText); // Output: hello%20world%21
This method encodes spaces as %20, making it ideal for query parameters.
For ASP.NET applications, you can also use HttpUtility.UrlEncode() (from System.Web), which encodes spaces as +:
using System.Web; string encodedText = HttpUtility.UrlEncode("hello world!"); Console.WriteLine(encodedText); // Output: hello+world%21
For .NET Core and later, Uri.EscapeDataString() is the preferred choice.
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.
Understanding the difference between COUNT() and COUNT(DISTINCT) in SQL is crucial for accurate data analysis.
COUNT() returns the total number of rows that match your query criteria, including duplicates, while COUNT(DISTINCT) returns the number of unique values in a specified column, effectively eliminating duplicates from the count.
For example, if you have a table of customer orders where a single customer can place multiple orders, COUNT(customer_id) would give you the total number of orders, whereas COUNT(DISTINCT customer_id) would tell you how many unique customers have placed orders.
The choice between these functions depends on your specific reporting needs. Use COUNT() when you need the total number of records, such as counting all sales transactions or total number of website visits.
Use COUNT(DISTINCT) when you need to know unique occurrences, like the number of different products sold or unique visitors to your website. It's also worth noting that COUNT(*) counts all rows including NULL values, while COUNT(column_name) excludes NULL values from that specific column, which can lead to different results depending on your data structure.
Example
-- Example table: customer_orders -- customer_id | order_date | product_id -- 1 | 2024-01-01 | 100 -- 1 | 2024-01-02 | 101 -- 2 | 2024-01-01 | 100 -- 3 | 2024-01-03 | 102 -- Count all orders SELECT COUNT(*) as total_orders FROM customer_orders; -- Result: 4 (counts all rows) -- Count unique customers who placed orders SELECT COUNT(DISTINCT customer_id) as unique_customers FROM customer_orders; -- Result: 3 (counts unique customer_ids: 1, 2, 3) -- Count unique products ordered SELECT COUNT(DISTINCT product_id) as unique_products FROM customer_orders; -- Result: 3 (counts unique product_ids: 100, 101, 102) -- Compare regular COUNT with COUNT DISTINCT SELECT COUNT(customer_id) as total_orders, COUNT(DISTINCT customer_id) as unique_customers FROM customer_orders; -- Result: total_orders = 4, unique_customers = 3
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